2. A Problem with Tubes
≪ 1. Week 1: Covering Compactness | Table of Contents | 3. Week 2: Compactness and Continuity ≫Embarassingly, I’ve already made a mistake: when proving that the product of compact spaces is compact, we said for each \(x\in X\), the “fibre” \(\left\lbrace x \right\rbrace\times Y\) admits a finite “fibrous subcover” \(V_x = U _{\alpha _1} \cup \cdots \cup U _{\alpha _n}\). I then said that if you intersect the projections of each of these sets onto the \(X\)-coordinate, this will give you the base of a tube; however, this is not correct, and I’d like to ammend that.
Lemma 1.
Let \(\left( X, d_X \right)\) and \(\left( Y, d_Y \right)\) be compact metric spaces. For any \(x\in X\), if \(V_x\) is an open subset of \(X\times Y\) containing \(\left\lbrace x \right\rbrace\times Y\), there is an open subset \(T_x\) of \(X\) such that \[\left\lbrace x \right\rbrace\times Y \subseteq T_x\times Y \subseteq V_x.\]
The proof proceeds in three steps:
- At every point \((x, y)\in \left\lbrace x \right\rbrace\times Y\), we can fit a small open square \((x, y) \in S_y\subseteq V_x\). Here, \[S_y = \left\lbrace \left( x’, y’ \right)\in X\times Y : d_X\left( x, x’ \right) < r_y\textrm{ and } d_Y \left( y, y’ \right) < r_y\right\rbrace = N _{r_y}(x) \times N _{r_y}(y)\] for some sufficiently small \(r_y > 0\). (This is a consequence of open-ness.)
- Since \(Y = \bigcup _{y\in Y} N _{r_y}(y)\), this is an open cover of \(Y\). By compactness of \(Y\), there exist finitely many \(y_1, \ldots, y_n\) such that \(Y = \bigcup _{j=1}^{n} N _{r _j}\left( y_j \right)\), where \(r_n = r _{y_n}\).
- Then, with \(y_n\) as above, we have \(\left\lbrace x \right\rbrace\times Y \subseteq \bigcup _{j=1}^{n} S _{y_j}\subseteq V_x\). If \(r = \min \left\lbrace r_1, \ldots, r_n \right\rbrace\), we have \[\left\lbrace x \right\rbrace\times Y \subseteq N_r(x) \times Y \subseteq \bigcup _{j=1}^{n} S _{y_n} \subseteq V_x.\] Take \(T_x = N_r(x)\).
In words, step 1 says to cover the fibre \(\left\lbrace x \right\rbrace\times Y\) by little squares that fit inside \(V_x\). Specifically, at each point \(y\), there is a square of “radius” \(r_y\) fitting inside \(V_x\), which we call \(S_y\).
Step 2 says to refine this to a cover of \(\left\lbrace x \right\rbrace\times Y\) by finitely many little squares, still fitting inside \(V_x\).
Step 3 says that all the little squares together must contain a thin rectangle, which is the “tube” we’re searching for, and it makes explicit how wide the rectangle must be.