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Tikhonov's Theorem Proof 📂Topology

Tikhonov's Theorem Proof

Theorem

An index set A\mathscr{A} is given.

If {Xα  αA}\left\{ X_{\alpha} \ | \ \alpha \in \mathscr{A} \right\} is a set of compact spaces, then X:=αAXα\displaystyle X : = \prod_{\alpha \in \mathscr{A}} X_{ \alpha} is compact.

Explanation

Though this theorem might seem like a trivial property at first glance, it’s actually the opposite. It appears trivial but is surprisingly difficult to prove, making it deserving of its own name. The fact that compactness is preserved under the Cartesian product of topological spaces is an exceptionally useful trait.

Proof

Let UαXαU_{\alpha} \subset X_{\alpha} be an open set of XαX_{\alpha}, and define the subbasis generating the product topology for the projection pα:XXαp_{\alpha} : X \to X_{\alpha} as follows: S:={pα1(Uα)  UαXα,αA} \mathscr{S} : = \left\{ p_{\alpha}^{-1} ( U_{\alpha} ) \ | \ U_{\alpha} \subset X_{\alpha} , \alpha \in \mathscr{A} \right\} To use Alexander’s subbase theorem, show that every open cover of S\mathscr{S} has a finite subcover.

Alexander’s Subbase Theorem: Suppose XX is a topological space. XX is compact. There exists some subbase S\mathscr{S} for XX such that every open cover of XX consisting of members of S\mathscr{S} has a finite subcover.


Part 1.

Assume there exists an open cover US\mathscr{U} \subset \mathscr{S} that does not have a finite subcover.

Let VV denote an open set from XαX_{\alpha}. For all αA\alpha \in \mathscr{A}: Uα={Vpα1(V)U} \mathscr{U}_{\alpha} = \left\{ V \mid p_{\alpha}^{-1} (V) \in \mathscr{U} \right\} then U\mathscr{U} is as follows: U=αA{pα1(V)  VUα} \mathscr{U} = \bigcup_{\alpha \in \mathscr{A}} \left\{ p_{\alpha}^{-1} (V) \ | \ V \in \mathscr{U}_{\alpha} \right\}


Part 2. Showing that Uα\mathscr{U}_{\alpha} is not an open cover of XαX_{\alpha}.

Assume that Uα\mathscr{U}_{\alpha} covers XαX_{\alpha}.

Since XαX_{\alpha} is compact, there exists some Uα1,,UαnUαU_{\alpha_{1}} , \cdots , U_{\alpha_{n}} \in \mathscr{U}_{\alpha} satisfying Xα=i=1nUαi\displaystyle X_{\alpha } = \bigcup_{i=1}^{n} U_{\alpha_{i}}. Then: {pα1(Uαi)  i=1,,n} \left\{ p_{\alpha}^{-1} ( U_{\alpha_{i}} ) \ | \ i = 1, \cdots , n \right\} becomes a finite subcover of U\mathscr{U}, which contradicts the definition of U\mathscr{U} in Part 1. Therefore, Uα\mathscr{U}_{\alpha} cannot cover XαX_{\alpha}.


Part 3.

Since Uα\mathscr{U}_{\alpha} is not a cover of XαX_{\alpha}, for all UαUαU_{\alpha} \in \mathscr{U}_{\alpha} there exists some xαXαx_{\alpha} \in X_{\alpha} satisfying xαUαx_{\alpha} \notin U_{\alpha}. Then for all UUU \in \mathscr{U}, pα1(xα)Up_{\alpha}^{-1} (x_{\alpha} ) \notin U, and U\mathscr{U} cannot cover XX. This means that there cannot exist an open cover US\mathscr{U} \subset \mathscr{S} that does not have a finite subcover. By Alexander’s subbase theorem, XX is compact.