-2
$\begingroup$

Let $X$ be a compact Hausdorff space. Suppose that for each $x \in X$, there is a neighborhood $U$ of $x$ and a positive integer $k$ such that $U$ can be imbedded in $\mathbb{R}^k$. Show that $X$ can be imbedded in $\mathbb{R}^N$ for some positive integer $N$.

$\endgroup$
  • 1
    $\begingroup$ It would be useful if you could tell us what you have tried. This helps us to understand what you already know so we can better focus our help. Also, people around here get touchy when people just post a question without motivation etc. as they are unwilling to simply do others homework for them, moreover they dislike it when a third party steps in and does the homework for them anyway so questions like this which merely states a question tend to get closed (pending adding more info). Adding in what you have tried and where you came across it will stop your question getting closed in this way. $\endgroup$ – user1729 Nov 3 '13 at 17:50
  • $\begingroup$ Grumpy Parsnip has indicated on Meta that zie wants this reopened. It appears the OP has abandoned it and is unlikely to return, so I'll concur. $\endgroup$ – dfeuer Nov 6 '13 at 6:27
3
$\begingroup$

I think this is the easiest way:

Fix a point $x$, and let $f \colon U_x \to (0,1)^k$ be an embedding of a neighborhood of $x$. Since $X$ is compact and Hausdorff, it is regular, so $x$ has an open neighborhood $W_x$ such that $\overline{W}_x \subseteq U_x$.

By compactness, $X$ can be covered by finitely many of the $W_x$'s - call them $W_1, \ldots, W_n$, and let $U_1, \ldots, U_n$ denote the corresponding $U_x$'s. Every compact Hausdorff space is normal, so by Urysohn's lemma there exist continuous functions $g_i \colon U_i \to \mathbb{R}$ such that $g_i = 1$ on $\overline W_i$ and $g_i = 0$ on the complement of $U_i$. Define

$$F(x) = (g_1(x)f_1(x), g_1(x), \ldots, g_n(x)f_n(x), g_n(x))$$

where $f_i \colon U_i \to \mathbb{R}^{k_i}$ is the embedding above. Thus $F$ defines a continuous map $F \colon X \to \mathbb{R}^{N}$ where

$$N = \sum_{i = 1}^n k_i + n$$

Moreover, $F$ is injective: if $x$ and $y$ are distinct points in $X$ then either $x$ and $y$ are both in the same $W_i$ or $x$ is in $W_i$ but $y$ is not in $W_i$. In the first case $g_i(x)f_i(x) \neq g_i(y)f_i(y)$ because $g_i(x) = g_i(y) = 1$ and $f_i$ is injective; in the second case $g_i(x) = 1$ while $g_i(y) < 1$.

I'd wager that the optimal number $N$ is smaller than the one that appears in this argument, but it does work.

$\endgroup$
  • $\begingroup$ How does invariance of domain apply? We don't know that it is an open embedding. $\endgroup$ – dfeuer Nov 3 '13 at 18:03
  • $\begingroup$ @dfeuer: That's a good point. However, I think the argument can be salvaged by defining the $V$'s and $W$'s to be the intersection of the $f(U)$'s with sufficiently small balls. I'll reflect on this before editing. $\endgroup$ – Paul Siegel Nov 3 '13 at 18:24
  • $\begingroup$ I have attempted to repair the problem I found. I don't know if I succeeded, or whether there are other problems remaining. Feel free to revert if I did something really stupid. $\endgroup$ – dfeuer Nov 3 '13 at 21:15
  • $\begingroup$ I believe the local embeddings need to be assumed bounded to ensure continuity of the products. I revised accordingly. $\endgroup$ – dfeuer Nov 3 '13 at 23:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.