Editing Banach–Alaoglu theorem (section)

===Elementary proof===

The following [[elementary proof]] does not utilize [[duality theory]] and requires only basic concepts from set theory, topology, and functional analysis. 
What is needed from topology is a working knowledge of [[Net (mathematics)|net]] [[Net convergence|convergence]] in [[topological space]]s and familiarity with the fact that a [[Continuous linear functional|linear functional is continuous]] if and only if it is [[Continuous linear operator#bounded on a neighborhood of a point|bounded on a neighborhood]] of the origin (see the articles on [[continuous linear functional]]s and [[sublinear functional]]s for details). 
Also required is a proper understanding of the technical details of how the space <math>\mathbb{K}^X</math> of all functions of the form <math>X \to \mathbb{K}</math> is identified as the [[Cartesian product]] <math display=inline>\prod_{x \in X} \mathbb{K},</math> and the relationship between [[pointwise convergence]], the [[product topology]], and [[Subspace topology|subspace topologies]] they induce on subsets such as the [[algebraic dual space]] <math>X^{\#}</math> and products of subspaces such as <math display=inline>\prod_{x \in X} B_{r_x}.</math> 
An explanation of these details is now given for readers who are interested. 

{{collapse top|title=Primer on product/function spaces, nets, and pointwise convergence|left=true}}

For every real <math>r,</math> <math>B_r ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{c \in \mathbb{K} : |c| \leq r\}</math> will denote the closed ball of radius <math>r</math> centered at <math>0</math> and <math>r U ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{r u : u \in U\}</math> for any <math>U \subseteq X,</math>

'''Identification of functions with tuples'''

The Cartesian product <math display=inline>\prod_{x \in X} \mathbb{K}</math> is usually thought of as the set of all <math>X</math>-indexed [[tuple]]s <math>s_{\bull} = \left(s_x\right)_{x \in X}</math> but, since tuples are technically just functions from an indexing set, it can also be identified with the space <math>\mathbb{K}^X</math> of all functions having prototype <math>X \to \mathbb{K},</math> as is now described: 

* {{em|Function <math>\to</math> Tuple}}: A function <math>s : X \to \mathbb{K}</math> belonging to <math>\mathbb{K}^X</math> is identified with its (<math>X</math>-indexed) "{{em|tuple of values}}" <math>s_{\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ (s(x))_{x \in X}.</math>
* {{em|Tuple <math>\to</math> Function}}: A tuple <math>s_{\bull} = \left(s_x\right)_{x \in X}</math> in <math display=inline>\prod_{x \in X} \mathbb{K}</math> is identified with the function <math>s : X \to \mathbb{K}</math> defined by <math>s(x) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ s_x</math>; this function's "tuple of values" is the original tuple <math>\left(s_x\right)_{x \in X}.</math>

This is the reason why many authors write, often without comment, the equality 
<math display=block>\mathbb{K}^X = \prod_{x \in X} \mathbb{K}</math>
and why the Cartesian product <math display=inline>\prod_{x \in X} \mathbb{K}</math> is sometimes taken as the definition of the set of maps <math>\mathbb{K}^X</math> (or conversely). 
However, the Cartesian product, being the [[Product (category theory)|(categorical) product]] in the [[Category (mathematics)|category]] of [[Category of sets|sets]] (which is a type of [[inverse limit]]), also comes equipped with associated maps that are known as its (coordinate) {{em|projections}}. 

The {{visible anchor|canonical projection of the Cartesian product}} at a given point <math>z \in X</math> is the function
<math display=block>\Pr{}_z : \prod_{x \in X} \mathbb{K} \to \mathbb{K} \quad \text{ defined by } \quad s_{\bull} = \left(s_x\right)_{x \in X} \mapsto s_z</math>
where under the above identification, <math>\Pr{}_z</math> sends a function <math>s : X \to \mathbb{K}</math> to
<math display=block>\Pr{}_z(s) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ s(z).</math> 
Stated in words, for a point <math>z</math> and function <math>s,</math> "plugging <math>z</math> into <math>s</math>" is the same as "plugging <math>s</math> into <math>\Pr{}_z</math>". 

In particular, suppose that <math>\left(r_x\right)_{x \in X}</math> are non-negative real numbers. 
Then <math>\prod_{x \in X} B_{r_x} \subseteq \prod_{x \in X} \mathbb{K} = \mathbb{K}^X,</math> where under the above identification of tuples with functions, <math>\prod_{x \in X} B_{r_x}</math> is the set of all functions <math>s \in \mathbb{K}^X</math> such that <math>s(x) \in B_{r_x}</math> for every <math>x \in X.</math> 

If a subset <math>U \subseteq X</math> [[Partition of a set|partitions]] <math>X</math> into <math>X = U \, \cup \,(X \setminus U)</math> then the linear bijection
<math display=block>\begin{alignat}{4}
H :\;&& \prod_{x \in X} \mathbb{K} &&\;\to    \;& \left(\prod_{u \in U} \mathbb{K}\right) \times \prod_{x \in X \setminus U} \mathbb{K} \\[0.3ex]
     && \left(f_x\right)_{x \in X} &&\;\mapsto\;& \left( \left(f_u\right)_{u \in U}, \; \left(f_x\right)_{x \in X \setminus U} \right) \\
\end{alignat}</math>
canonically identifies these two Cartesian products; moreover, this map is a [[homeomorphism]] when these products are endowed with their product topologies. 
In terms of function spaces, this bijection could be expressed as 
<math display=block>\begin{alignat}{4}
H :\;&& \mathbb{K}^X &&\;\to    \;& \mathbb{K}^U \times \mathbb{K}^{X \setminus U} \\[0.3ex]
     && f &&\;\mapsto\;& \left(f\big\vert_U, \; f\big\vert_{X \setminus U}\right) \\
\end{alignat}.</math>

'''Notation for nets and function composition with nets'''

A [[Net (mathematics)|net]] <math>x_{\bull} = \left(x_i\right)_{i \in I}</math> in <math>X</math> is by definition a function <math>x_{\bull} : I \to X</math> from a non-empty [[directed set]] <math>(I, \leq).</math> 
Every [[sequence]] in <math>X,</math> which by definition is just a function of the form <math>\N \to X,</math> is also a net. 
As with sequences, the value of a net <math>x_{\bull}</math> at an index <math>i \in I</math> is denoted by <math>x_i</math>; however, for this proof, this value <math>x_i</math> may also be denoted by the usual function parentheses notation <math>x_{\bull}(i).</math> 
Similarly for [[function composition]], if <math>F : X \to Y</math> is any function then the net (or sequence) that results from "plugging <math>x_{\bull}</math> into <math>F</math>" is just the function <math>F \circ x_{\bull} : I \to Y,</math> although this is typically denoted by <math>\left(F\left(x_i\right)\right)_{i \in I}</math> (or by <math>\left(F\left(x_i\right)\right)_{i=1}^{\infty}</math> if <math>x_{\bull}</math> is a sequence). 
In the proofs below, this resulting net may be denoted by any of the following notations 
<math display=block>F\left(x_{\bull}\right) = \left(F\left(x_i\right)\right)_{i \in I} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ F \circ x_{\bull},</math> 
depending on whichever notation is cleanest or most clearly communicates the intended information. 
In particular, if <math>F : X \to Y</math> is continuous and <math>x_{\bull} \to x</math> in <math>X,</math> then the conclusion commonly written as <math>\left(F\left(x_i\right)\right)_{i \in I} \to F(x)</math> may instead be written as <math>F\left(x_{\bull}\right) \to F(x)</math> or <math>F \circ x_{\bull} \to F(x).</math> 

'''Topology'''

The set <math display=inline>\mathbb{K}^X = \prod_{x \in X} \mathbb{K}</math> is assumed to be endowed with the [[product topology]]. It is well known that the product topology is identical to the [[topology of pointwise convergence]]. 
This is because given <math>f</math> and a [[Net (mathematics)|net]] <math>\left(f_i\right)_{i \in I},</math> where <math>f</math> and every <math>f_i</math> is an element of <math display=inline>\mathbb{K}^X = \prod_{x \in X} \mathbb{K},</math> then the net <math>\left(f_i\right)_{i \in I} \to f</math> [[Convergent net|converges]] in the product topology if and only if

:for every <math>z \in X,</math> the net <math>\Pr{}_z\left(\left(f_i\right)_{i \in I}\right) \to \Pr{}_z(f)</math> converges in <math>\mathbb{K},</math> 

where because <math>\;\Pr{}_z(f) = f(z)\;</math> and 
<math display=inline>\Pr{}_z\left(\left(f_i\right)_{i \in I}\right) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left(\Pr{}_z\left(f_i\right)\right)_{i \in I} = \left(f_i(z)\right)_{i \in I},</math> 
this happens if and only if 

:for every <math>z \in X,</math> the net <math>\left(f_i(z)\right)_{i \in I} \to f(z)</math> converges in <math>\mathbb{K},</math> 

Thus <math>\left(f_i\right)_{i \in I}</math> converges to <math>f</math> in the product topology if and only if it converges to <math>f</math> pointwise on <math>X.</math>

This proof will also use the fact that the topology of pointwise convergence is preserved when passing to [[topological subspace]]s. 
This means, for example, that if for every <math>x \in X,</math> <math>S_x \subseteq \mathbb{K}</math> is some [[Topological subspace|(topological) subspace]] of <math>\mathbb{K}</math> then the topology of pointwise convergence (or equivalently, the product topology) on <math display=inline>\prod_{x \in X} S_x</math> is equal to the [[subspace topology]] that the set <math display=inline>\prod_{x \in X} S_x</math> inherits from <math display=inline>\prod_{x \in X} \mathbb{K}.</math> 
And if <math>S_x</math> is closed in <math>\mathbb{K}</math> for every <math>x \in X,</math> then <math display=inline>\prod_{x \in X} S_x</math> is a closed subset of <math display=inline>\prod_{x \in X} \mathbb{K}.</math> 

'''Characterization of <math>\sup_{u \in U} |f(u)| \leq r</math>'''

An important fact used by the proof is that for any real <math>r,</math> 
<math display=block>\sup_{u \in U} |f(u)| \leq r \qquad \text{ if and only if } \qquad f(U) \subseteq B_r</math>
where <math>\,\sup\,</math> denotes the [[supremum]] and <math>f(U) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{f(u) : u \in U\}.</math> 
As a side note, this characterization does not hold if the closed ball <math>B_r</math> is replaced with the open ball <math>\{c \in \mathbb{K} : |c| < r\}</math> (and replacing <math>\;\sup_{u \in U} |f(u)| \leq r\;</math> with the strict inequality <math>\;\sup_{u \in U} |f(u)| < r\;</math> will not change this; for counter-examples, consider <math>X ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \mathbb{K}</math> and the [[identity map]] <math>f ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \operatorname{Id}</math> on <math>X</math>). 
{{collapse bottom}} 

The essence of the Banach–Alaoglu theorem can be found in the next proposition, from which the Banach–Alaoglu theorem follows. 
Unlike the Banach–Alaoglu theorem, this proposition does {{em|not}} require the vector space <math>X</math> to endowed with any topology. 

{{Math theorem|name=Proposition{{sfn|Narici|Beckenstein|2011|pp=225-273}}|math_statement=
Let <math>U</math> be a subset of a vector space <math>X</math> over the field <math>\mathbb{K}</math> (where <math>\mathbb{K} = \R \text{ or } \mathbb{K} = \Complex</math>) and for every real number <math>r,</math> endow the closed ball <math display=inline>B_r ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{s \in \mathbb{K} : |s| \leq r\}</math> with its [[Subspace topology|usual topology]] (<math>X</math> need not be endowed with any topology, but <math>\mathbb{K}</math> has its usual [[Euclidean topology]]). 
Define
<math display=block>U^{\#} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \Big\{f \in X^{\#} ~:~ \sup_{u \in U} |f(u)| \leq 1\Big\}.</math>

If for every <math>x \in X,</math> <math>r_x > 0</math> is a real number such that <math>x \in r_x U,</math> then <math>U^{\#}</math> is a closed and [[Compact space|compact]] [[Topological subspace|subspace]] of the [[product space]] <math>\prod_{x \in X} B_{r_x}</math> (where because this [[product topology]] is identical to the [[topology of pointwise convergence]], which is also called the [[weak-* topology]] in functional analysis, this means that <math>U^{\#}</math> is compact in the weak-* topology or "weak-* compact" for short).
}}

Before proving the proposition above, it is first shown how the Banach–Alaoglu theorem follows from it (unlike the proposition, Banach–Alaoglu assumes that <math>X</math> is a [[topological vector space]] (TVS) and that <math>U</math> is a neighborhood of the origin). 

{{Math proof|drop=hidden|title=Proof that Banach–Alaoglu follows from the proposition above|proof=
Assume that <math>X</math> is a [[topological vector space]] with continuous dual space <math>X^{\prime}</math> and that <math>U</math> is a neighborhood of the origin. 
Because <math>U</math> is a neighborhood of the origin in <math>X,</math> it is also an [[Absorbing set|absorbing subset]] of <math>X,</math> so for every <math>x \in X,</math> there exists a real number <math>r_x > 0</math> such that <math>x \in r_x U.</math> 
Thus the hypotheses of the above proposition are satisfied, and so the set <math>U^{\#}</math> is therefore compact in the [[weak-* topology]].  
The proof of the Banach–Alaoglu theorem will be complete once it is shown that <math>U^{\#} = U^{\circ},</math><ref group=note>If <math>\tau</math> denotes the topology that <math>X</math> is (originally) endowed with, then the equality <math>U^{\circ} = U^{\#}</math> shows that the polar <math>U^{\circ} = \Big\{f \in X^{\prime} ~:~ \sup_{u \in U} |f(u)| \leq 1\Big\}</math> of <math>U</math> is dependent {{em|only}} on <math>U</math> (and <math>X^{\#}</math>) and that the rest of the topology <math>\tau</math> can be ignored.  To clarify what is meant, suppose <math>\sigma</math> is any TVS topology on <math>X</math> such that the set <math>U</math> is (also) a neighborhood of the origin in <math>(X, \sigma).</math>  Denote the continuous dual space of <math>(X, \sigma)</math> by <math>(X, \sigma)^{\prime}</math> and denote the polar of <math>U</math> with respect to <math>(X, \sigma)</math> by 
<math display=block>U^{\circ, \sigma} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \Big\{f \in (X, \sigma)^{\prime} ~:~ \sup_{u \in U} |f(u)| \leq 1\Big\}</math>
so that <math>U^{\circ, \tau}</math> is just the set <math>U^{\circ}</math> from above. 
Then <math>U^{\circ, \tau} = U^{\circ, \sigma}</math> because both of these sets are equal to <math>U^{\#}.</math>  Said differently, the polar set <math>U^{\circ, \sigma}</math>'s defining "requirement" that <math>U^{\circ, \sigma}</math> be a subset of the '''{{em|continuous}}''' dual space <math>(X, \sigma)^{\prime}</math> is inconsequential and can be ignored because it does not have any effect on the resulting set of linear functionals. However, if <math>\nu</math> is a TVS topology on <math>X</math> such that <math>U</math> is {{em|not}} a neighborhood of the origin in <math>(X, \nu)</math> then the polar <math>U^{\circ, \nu}</math> of <math>U</math> with respect to <math>(X, \nu)</math> is not guaranteed to equal <math>U^{\#}</math> and so the topology <math>\nu</math> can not be ignored.</ref> 
where recall that <math>U^{\circ}</math> was defined as
<math display=block>U^{\circ} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \Big\{f \in X^{\prime} ~:~ \sup_{u \in U} |f(u)| \leq 1\Big\} ~=~ U^{\#} \cap X^{\prime}.</math>

'''Proof that <math>U^{\circ} = U^{\#}:</math>''' 
Because <math>U^{\circ} = U^{\#} \cap X^{\prime},</math> the conclusion is equivalent to <math>U^{\#} \subseteq X^{\prime}.</math> 
If <math>f \in U^{\#}</math> then <math>\;\sup_{u \in U} |f(u)| \leq 1,\,</math> which states exactly that the linear functional <math>f</math> is bounded on the neighborhood <math>U;</math> thus <math>f</math> is a [[continuous linear functional]] (that is, <math>f \in X^{\prime}</math>), as desired. 
<math>\blacksquare</math>
}}

{{Math proof|drop=hidden|title=Proof of Proposition|proof=
The [[Product space (topology)|product space]] <math display=inline>\prod_{x \in X} B_{r_x}</math> is compact by [[Tychonoff's theorem]] (since each closed ball <math>B_{r_x} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{s \in \mathbb{K} : |s| \leq r_x\}</math> is a [[Hausdorff space|Hausdorff]]<ref group=note>Because every <math>B_{r_x}</math> is also a [[Hausdorff space]], the conclusion that <math>\prod_{x \in X} B_{r_x}</math> is compact only requires the so-called "Tychonoff's theorem for compact Hausdorff spaces," which is equivalent to the [[ultrafilter lemma]] and strictly weaker than the [[axiom of choice]].</ref> [[compact space]]). Because a closed subset of a compact space is compact, the proof of the proposition will be complete once it is shown that
<math display=block>U^{\#} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \Big\{f \in X^{\#} ~:~ \sup_{u \in U} |f(u)| \leq 1\Big\} ~=~ \left\{f \in X^{\#} ~:~ f(U) \subseteq B_1\right\}</math>
is a closed subset of <math display=inline>\prod_{x \in X} B_{r_x}.</math> 
The following statements guarantee this conclusion:
#<math>U^{\#} \subseteq \prod_{x \in X} B_{r_x}.</math>
#<math>U^{\#}</math> is a closed subset of the [[product space]] <math>\prod_{x \in X} \mathbb{K} = \mathbb{K}^X.</math>

'''Proof of (1)''': 

For any <math>z \in X,</math> let <math display=inline>\Pr{}_z : \prod_{x \in X} \mathbb{K} \to \mathbb{K}</math> denote the projection to the <math>z</math><sup>th</sup> coordinate ([[#canonical projection of the Cartesian product|as defined above]]). 
To prove that <math display=inline>U^{\#} \subseteq \prod_{x \in X} B_{r_x},</math> it is sufficient (and necessary) to show that <math>\Pr{}_x\left(U^{\#}\right) \subseteq B_{r_x}</math> for every <math>x \in X.</math> 
So fix <math>x \in X</math> and let <math>f \in U^{\#}.</math> 
Because <math>\Pr{}_x(f) \,=\, f(x),</math> it remains to show that <math>f(x) \in B_{r_x}.</math> 
Recall that <math>r_x > 0</math> was defined in the proposition's statement as being any positive real number that satisfies <math>x \in r_x U</math> (so for example, <math>r_u := 1</math> would be a valid choice for each <math>u \in U</math>), which implies <math>\,u_x ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \frac{1}{r_x} \, x \in U.\,</math> 
Because <math>f</math> is a [[Strict positive homogeneity|positive homogeneous]] function that satisfies <math>\;\sup_{u \in U} |f(u)| \leq 1,\,</math> 
<math display=block>\frac{1}{r_x}|f(x)|
= \left|\frac{1}{r_x} f(x)\right|
= \left|f\left(\frac{1}{r_x} x\right)\right|
= \left|f\left(u_x\right)\right| 
\leq \sup_{u \in U} |f(u)| 
\leq 1.</math> 

Thus <math>|f(x)| \leq r_x,</math> which shows that <math>f(x) \in B_{r_x},</math> as desired. 

'''Proof of (2)''': 

The [[algebraic dual space]] <math>X^{\#}</math> is always a closed subset of <math display=inline>\mathbb{K}^X = \prod_{x \in X} \mathbb{K}</math> (this is proved in [[#Algebraic dual space is closed in the space of all functions|the lemma below]] for readers who are not familiar with this result). 
The set
<math display=block>\begin{alignat}{9}
U_{B_1} 
&\,\stackrel{\scriptscriptstyle\text{def}}{=}\, \Big\{ ~~\;~~\;~~\;~~ f\ \in \mathbb{K}^X  ~~\;~~ : \sup_{u \in U} |f(u)| \leq 1\Big\} \\
&= \big\{ ~~\;~~\;~~\;~~f \, \in \mathbb{K}^X  ~~\;~~ : f(u) \in B_1 \text{ for all } u \in U\big\} \\
&= \Big\{\left(f_x\right)_{x \in X} \in \prod_{x \in X} \mathbb{K} \,~:~ \; ~f_u~ \in B_1  \text{ for all } u \in U\Big\} \\
&= \prod_{x \in X} C_x \quad \text{ where } \quad C_x ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \begin{cases}
B_1        & \text{ if } x \in U \\
\mathbb{K} & \text{ if } x \not\in U \\
\end{cases} \\
\end{alignat}</math>
is closed in the [[product topology]] on <math>\prod_{x \in X} \mathbb{K} = \mathbb{K}^X</math> since it is a product of closed subsets of <math>\mathbb{K}.</math> 
Thus <math>U_{B_1} \cap X^{\#} = U^{\#}</math> is an intersection of two closed subsets of <math>\mathbb{K}^X,</math> which proves (2).<ref group=note>The conclusion <math>U_{B_1} = \prod_{x \in X} C_x</math> can be written as <math>U_{B_1} ~=~ \Big(\prod_{u \in U} B_1\Big) \times \prod_{x \in X \setminus U} \mathbb{K}.</math> The set <math>U^{\#}</math> may thus equivalently be defined by <math>U^{\#} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ X^{\#} \cap \left[\Big(\prod_{u \in U} B_1\Big) \times \prod_{x \in X \setminus U} \mathbb{K}\right].</math> Rewriting the definition in this way helps make it apparent that the set <math>U^{\#}</math> is closed in <math>\prod_{x \in X} \mathbb{K}</math> because [[#Algebraic dual space is closed in the space of all functions|this is true of <math>X^{\#}.</math>]]</ref>
<math>\blacksquare</math>
}}

The conclusion that the set <math>U_{B_1} = \left\{f \in \mathbb{K}^X : f(U) \subseteq B_1\right\}</math> is closed can also be reached by applying the following more general result, this time proved using nets, to the special case <math>Y := \mathbb{K}</math> and <math>B := B_1.</math><!-- NOTE: This added generality does not make the proof more difficult because the proof is nearly identical to the proof of the special space where Y = K and B = B_1. In fact, the generality simplified the proof slightly (compared to the older version from e.g. 26 November 2021) -->

:'''Observation''': If <math>U \subseteq X</math> is any set and if <math>B \subseteq Y</math> is a [[Closed set|closed]] subset of a topological space <math>Y,</math> then <math>U_B ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left\{f \in Y^X : f(U) \subseteq B\right\}</math> is a closed subset of <math>Y^X</math> in the topology of pointwise convergence. 

:'''Proof of observation''': Let <math>f \in Y^X</math> and suppose that <math>\left(f_i\right)_{i \in I}</math> is a net in <math>U_B</math> that converges pointwise to <math>f.</math> It remains to show that <math>f \in U_B,</math> which by definition means <math>f(U) \subseteq B.</math> For any <math>u \in U,</math> because <math>\left(f_i(u)\right)_{i \in I} \to f(u)</math> in <math>Y</math> and every value <math>f_i(u) \in f_i(U) \subseteq B</math> belongs to the closed (in <math>Y</math>) subset <math>B,</math> so too must this net's limit belong to this closed set; thus <math>f(u) \in B,</math> which completes the proof. <math>\blacksquare</math>

{{math theorem|name=Lemma (<math>X^{\#}</math> is closed in <math>\mathbb{K}^X</math>){{anchor|Algebraic dual space is closed in the space of all functions}}|note=|style=|math_statement= 
The [[algebraic dual space]] <math>X^{\#}</math> of any vector space <math>X</math> over a field <math>\mathbb{K}</math> (where <math>\mathbb{K}</math> is <math>\R</math> or <math>\Complex</math>) is a closed subset of <math display=inline>\mathbb{K}^X = \prod_{x \in X} \mathbb{K}</math> in the topology of pointwise convergence. (The vector space <math>X</math> need not be endowed with any topology). 
}}

{{collapse top|title=Proof of lemma|left=true}}
Let <math>f \in \mathbb{K}^X</math> and suppose that <math>f_{\bull} = \left(f_i\right)_{i \in I}</math> is a net in <math>X^{\#}</math> the converges to <math>f</math> in <math>\mathbb{K}^X.</math> 
To conclude that <math>f \in X^{\#},</math> it must be shown that <math>f</math> is a linear functional. So let <math>s</math> be a scalar and let <math>x, y \in X.</math> 

For any <math>z \in X,</math> let <math>f_{\bull}(z) : I \to \mathbb{K}</math> denote {{em|<math>f_{\bull}</math>'s net of values at <math>z</math>}}
<math display=block>f_{\bull}(z) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left(f_i(z)\right)_{i \in I}.</math>
Because <math>f_{\bull} \to f</math> in <math>\mathbb{K}^X,</math> which has the topology of pointwise convergence, <math>f_{\bull}(z) \to f(z)</math> in <math>\mathbb{K}</math> for every <math>z \in X.</math> 
By using <math>x, y, sx, \text{ and } x + y,</math> in place of <math>z,</math> it follows that each of the following nets of scalars converges in <math>\mathbb{K}:</math>
<math display=block>f_{\bull}(x) \to f(x), \quad f_{\bull}(y) \to f(y), \quad f_{\bull}(x + y) \to f(x + y), \quad \text{ and } \quad f_{\bull}(sx) \to f(sx).</math>


Proof that <math>f(s x) = s f(x):</math> 
Let <math>M : \mathbb{K} \to \mathbb{K}</math> be the "multiplication by <math>s</math>" map defined by <math>M(c) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ s c.</math> 
Because <math>M</math> is continuous and <math>f_{\bull}(x) \to f(x)</math> in <math>\mathbb{K},</math> it follows that <math>M\left(f_{\bull}(x)\right) \to M(f(x))</math> where the right hand side is <math>M(f(x)) = s f(x)</math> and the left hand side is
<math display=block>\begin{alignat}{4}
M\left(f_{\bull}(x)\right) 
\stackrel{\scriptscriptstyle\text{def}}{=}&~ M \circ f_{\bull}(x)                            && \text{ by definition of notation } \\
=&~ \left(M\left(f_i(x)\right)\right)_{i \in I}  ~~~ && \text{ because } f_{\bull}(x) = \left(f_i(x)\right)_{i \in I} : I \to \mathbb{K} \\
=&~ \left(s f_i(x)\right)_{i \in I}                  && M\left(f_i(x)\right) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ s f_i(x) \\
=&~ \left(f_i(s x)\right)_{i \in I}                  && \text{ by linearity of } f_i \\
=&~ f_{\bull}(sx)                                    && \text{ notation }
\end{alignat}</math>
which proves that <math>f_{\bull}(sx) \to s f(x).</math> Because also <math>f_{\bull}(sx) \to f(sx)</math> and limits in <math>\mathbb{K}</math> are unique, it follows that <math>s f(x) = f(s x),</math> as desired. 


Proof that <math>f(x + y) = f(x) + f(y):</math> 
Define a net <math>z_{\bull} = \left(z_i\right)_{i \in I} : I \to \mathbb{K} \times \mathbb{K}</math> by letting <math>z_i ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left(f_i(x), f_i(y)\right)</math> for every <math>i \in I.</math> 
Because <math>f_{\bull}(x) = \left(f_i(x)\right)_{i \in I} \to f(x)</math> and <math>f_{\bull}(y) = \left(f_i(y)\right)_{i \in I} \to f(y),</math> it follows that <math>z_{\bull} \to ( f(x), f(y) )</math> in <math>\mathbb{K} \times \mathbb{K}.</math> 
Let <math>A : \mathbb{K} \times \mathbb{K} \to \mathbb{K}</math> be the addition map defined by <math>A(x, y) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ x + y.</math> 
The continuity of <math>A</math> implies that <math>A\left(z_{\bull}\right) \to A(f(x), f(y))</math> in <math>\mathbb{K}</math> where the right hand side is <math>A(f(x), f(y)) = f(x) + f(y)</math> and the left hand side is
<math display=block>A\left(z_{\bull}\right) ~\stackrel{\scriptscriptstyle\text{def}}{=}~ A \circ z_{\bull} 
= \left(A\left(z_i\right)\right)_{i \in I} 
= \left(A\left(f_i(x), f_i(y)\right)\right)_{i \in I} 
= \left(f_i(x) + f_i(y)\right)_{i \in I} 
= \left(f_i(x + y)\right)_{i \in I} 
= f_{\bull}(x + y)
</math>
which proves that <math>f_{\bull}(x + y)  \to f(x) + f(y).</math> Because also <math>f_{\bull}(x + y) \to f(x + y),</math> it follows that <math>f(x + y) = f(x) + f(y),</math> as desired. 
<math>\blacksquare</math>

<!-- Removed information
Alternative proof outline: 
The proof above did not involve [[Cauchy net]]s or even anything related to the [[uniform structure]] (specifically, the [[product uniformity]]) that the product space <math>\prod_{x \in X} \mathbb{K} = \mathbb{K}^X</math> is endowed with. 
However, it is possible to prove this lemma by first showing that <math>X^{\#}</math> is a [[Complete topological vector space|complete TVS]] and then concluding that it is a closed subset of the Hausdorff space <math>\mathbb{K}^X</math> (because every complete subspace of a Hausdorff space is necessarily a closed subset). 
This involves dealing with [[Cauchy net]]s, where it can be shown that a net <math>f_{\bull}</math> is Cauchy in the topology of pointwise convergence if and only if for every point <math>x \in X,</math> the net of values <math>\left(f_i(x)\right)_{i \in I}</math> is Cauchy in <math>\mathbb{K}.</math> 
Showing that <math>X^{\#}</math> is a complete space requires showing that every Cauchy net <math>f_{\bull}</math> in <math>X^{\#}</math> converges to some element <math>f</math> of <math>X^{\#}.</math> 
However, such a proof would require additional steps such as constructing the function <math>f</math> from the net <math>f_{\bull}</math> and then proving that <math>f \in X^{\#}</math> and that <math>f_{\bull} \to f</math> in <math>X^{\#}.</math> 
--- End: Removed information -->
{{collapse bottom}}<!-- END OF PROOF OF LEMMA -->

[[#Algebraic dual space is closed in the space of all functions|The lemma above]] actually also follows from its corollary below since <math>\prod_{x \in X} \mathbb{K}</math> is a Hausdorff [[complete uniform space]] and any subset of such a space (in particular <math>X^{\#}</math>) is closed if and only if it is complete. 

{{math theorem|name=Corollary to lemma (<math>X^{\#}</math> is weak-* complete)|note=|style=|math_statement= 
When the [[algebraic dual space]] <math>X^{\#}</math> of a vector space <math>X</math> is equipped with the topology <math>\sigma\left(X^{\#}, X\right)</math> of pointwise convergence (also known as the weak-* topology) then the resulting [[topological space]] <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right)</math> is a [[Complete topological vector space|complete]] [[Hausdorff space|Hausdorff]] [[Locally convex topological vector space|locally convex]] [[topological vector space]]. 
}}

{{collapse top|title=Proof of corollary to lemma|left=true}}
<!-- {{math proof|title=Proof of corollary|proof=}} -->
Because the underlying field <math>\mathbb{K}</math> is a complete Hausdorff locally convex topological vector space, the same is true of the [[product space]] <math display=inline>\mathbb{K}^X = \prod_{x \in X} \mathbb{K}.</math> 
A closed subset of a complete space is complete, so by the lemma, the space <math>\left(X^{\#}, \sigma\left(X^{\#}, X\right)\right)</math> is complete.
<math>\blacksquare</math> 
{{collapse bottom}}


The above elementary proof of the Banach–Alaoglu theorem actually shows that if <math>U \subseteq X</math> is any subset that satisfies <math>X = (0, \infty) U ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \{r u : r > 0, u \in U\}</math> (such as any [[Absorbing set|absorbing subset]] of <math>X</math>), then <math>U^{\#} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left\{f \in X^{\#} : f(U) \subseteq B_1\right\}</math> is a [[Weak-* topology|weak-* compact]] subset of <math>X^{\#}.</math> 

As a side note, with the help of the above elementary proof, it may be shown (see this footnote)<ref group=proof>
<!-- It suffices to prove that <math>P = \left(\cap \operatorname{Box}_P\right) \cap X^{\#} </math> because then <math>P \subseteq X^{\prime} \subseteq X^{\#}</math> will imply <math>P = \left(\cap \operatorname{Box}_P\right) \cap X^{\prime}.</math> -->
For any non-empty subset <math>A \subseteq [0, \infty),</math> the equality <math>\cap \left\{B_a : a \in A\right\} = B_{\inf_{} A}</math> holds (the intersection on the left is a closed, rather than open, disk − possibly of radius <math>0</math> − because it is an intersection of closed subsets of <math>\mathbb{K}</math> and so must itself be closed). For every <math>x \in X,</math> let <math>m_x = \inf_{} \left\{ R_x : R_{\bull} \in T_P \right\}</math> so that the previous set equality implies <math>\cap \operatorname{Box}_P = \bigcap_{R_{\bull} \in T_P} \prod_{x \in X} B_{R_x} = \prod_{x \in X} \bigcap_{R_{\bull} \in T_P} B_{R_x} = \prod_{x \in X} B_{m_x}.</math> From <math>P \subseteq \cap \operatorname{Box}_P</math> it follows that <math>m_{\bull} \in T_P</math> and <math>\cap \operatorname{Box}_P \in \operatorname{Box}_P,</math> thereby making <math>\cap \operatorname{Box}_P</math> the [[least element]] of <math>\operatorname{Box}_P</math> with respect to <math>\,\subseteq.\,</math> (In fact, the [[Family of sets|family]] <math>\operatorname{Box}_P</math> is closed under (non-[[Nullary intersection|nullary]]) arbitrary intersections and also under finite unions of at least one set). The elementary proof showed that <math>T_P</math> and <math>\operatorname{Box}_P</math> are not empty and moreover, it also even showed that <math>T_P</math> has an element <math>\left(r_x\right)_{x \in X}</math> that satisfies <math>r_u = 1</math> for every <math>u \in U,</math> which implies that <math>m_u \leq 1</math> for every <math>u \in U.</math> The inclusion <math>P ~\subseteq~ \left(\cap \operatorname{Box}_P\right) \cap X^{\prime} ~\subseteq~ \left(\cap \operatorname{Box}_P\right) \cap X^{\#}</math> is immediate; to prove the reverse inclusion, let <math>f \in \left(\cap \operatorname{Box}_P\right) \cap X^{\#}.</math> By definition, <math>f \in P ~\stackrel{\scriptscriptstyle\text{def}}{=}~ U^{\#}</math> if and only if <math>\sup_{u \in U} |f(u)| \leq 1,</math> so let <math>u \in U</math> and it remains to show that <math>|f(u)| \leq 1.</math> From <math>f \in \cap \operatorname{Box}_P = \prod B_{m_\bull},</math> it follows that <math>f(u) = \Pr{}_u (f) \in \Pr{}_u \left(\prod_{x \in X} B_{m_x}\right) = B_{m_u},</math> which implies that <math>|f(u)| \leq m_u \leq 1,</math> as desired. 
<math>\blacksquare</math>
</ref> 
that there exist <math>X</math>-indexed non-negative real numbers <math>m_{\bull} = \left(m_x\right)_{x \in X}</math> such that 
<math display=block>\begin{alignat}{4}
U^{\circ} 
&= U^{\#}     && \\
&= X^{\#}     && \cap \prod_{x \in X} B_{m_x} \\
&= X^{\prime} && \cap \prod_{x \in X} B_{m_x} \\
\end{alignat}</math>
where these real numbers <math>m_{\bull}</math> can also be chosen to be "minimal" in the following sense: 
using <math>P ~\stackrel{\scriptscriptstyle\text{def}}{=}~ U^{\circ}</math> (so <math>P = U^{\#}</math> as in the proof) and defining the notation <math>\prod B_{R_\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \prod_{x \in X} B_{R_x}</math> for any <math>R_{\bull} = \left(R_x\right)_{x \in X} \in \R^X,</math> if
<math display=block>T_P ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left\{R_{\bull} \in \R^X ~:~ P \subseteq \prod B_{R_\bull}\right\}</math> 
then <math>m_{\bull} \in T_P</math> and for every <math>x \in X,</math> <math>m_x = \inf \left\{ R_x : R_{\bull} \in T_P \right\},</math> 
which shows that these numbers <math>m_{\bull}</math> are unique; indeed, this [[infimum]] formula can be used to define them. 

In fact, if <math>\operatorname{Box}_P</math> denotes the set of all such products of closed balls containing the polar set <math>P,</math>
<math display=block>\operatorname{Box}_P ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left\{ \prod B_{R_\bull} ~:~ R_{\bull} \in T_P \right\} ~=~ \left\{ \prod B_{R_\bull} ~:~ P \subseteq \prod B_{R_\bull} \right\},</math> 
then 
<math display=inline>\prod B_{m_\bull} = \cap \operatorname{Box}_P \in \operatorname{Box}_P</math> 
where <math display=inline>\bigcap \operatorname{Box}_P</math> denotes the intersection of all sets belonging to <math>\operatorname{Box}_P.</math>

This implies (among other things<ref group=note>This tuple <math>m_{\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left(m_x\right)_{x \in X}</math> is the [[least element]] of <math>T_P</math> with respect to natural induced pointwise [[partial order]] defined by <math>R_{\bull} \leq S_{\bull}</math> if and only if <math>R_x \leq S_x</math> for every <math>x \in X.</math> Thus, every neighborhood <math>U</math> of the origin in <math>X</math> can be associated with this unique (minimum) function <math>m_{\bull} : X \to [0, \infty).</math> For any <math>x \in X,</math> if <math>r > 0</math> is such that <math>x \in r U</math> then <math>m_x \leq r</math> so that in particular, <math>m_0 = 0</math> and <math>m_u \leq 1</math> for every <math>u \in U.</math></ref>) 
that <math display=inline>\prod B_{m_\bull} = \prod_{x \in X} B_{m_x}</math> the unique [[least element]] of <math>\operatorname{Box}_P</math> with respect to <math>\,\subseteq;</math> this may be used as an alternative definition of this (necessarily [[Convex set|convex]] and [[Balanced set|balanced]]) set.  
The function <math>m_{\bull} ~\stackrel{\scriptscriptstyle\text{def}}{=}~ \left(m_x\right)_{x \in X} : X \to [0, \infty)</math> is a [[seminorm]] and it is unchanged if <math>U</math> is replaced by the [[convex balanced hull]] of <math>U</math> (because <math>U^{\#} = [\operatorname{cobal} U]^{\#}</math>). 
Similarly, because <math>U^{\circ} = \left[\operatorname{cl}_X U\right]^{\circ},</math> <math>m_{\bull}</math> is also unchanged if <math>U</math> is replaced by its [[Closure (topology)|closure]] in <math>X.</math>