I'm really struggling trying to comprehend some aspects of embeddings. In particular, the problem I'm having is the relation between embeddings and homeomorphisms.
Let $C$ be a compact Hausdorff space with the property that for all $c \in C$, we can find a neigborhood (open) that can be embedded in $\mathbb{R}^{k}$ for some k. Does this mean that all open neigborhoods are homeomorphic to $\mathbb{R}^{k}$? Can I call this space $C$ as a $k$-manifold?
I want to prove that $C$ in fact can be embedded in some $\mathbb{R}^{N}$ by using the embedding theorem for compact manifolds, but to use it it I have to be able to show that C is in fact a manifold, thats why I want to see if the neigborhoods are all homeomorphic to $\mathbb{R}^{k}$.
Any comments, proofs, resources are welcome.
Best Answer
No. Any compact $C \subset \mathbb R^k$ has the property that each $c \in C$ has an open neigborhood in $C$ that can be embedded in $\mathbb R^k$. Actually we may take any open neigborhood $U$ of $c$ in $C$ since $U \subset \mathbb R^k$.
I think you mean that each $c \in C$ has an open neighborhood which can embedded as an open subset of $\mathbb R^k$. Then in fact $C$ is a $k$-manifold. But of course you cannot expect that that all open neigborhoods are homeomorphic to $\mathbb R^k$. There are such "nice" neighborhoods, but also other neigborhoods (just remove a compact subset from a nice neighborhood).
Update on request:
Let $C$ be a compact space such that for each $c \in C$ there exists an open neighborhood $U_c \subset C$ and an embedding $\phi_c : U_c \to \mathbb R^{k(c)}$ for some $k(c)$. We shall construct an embedding $h : C \to \mathbb R^N$ for a sufficiently large $N$. Since $C$ is compact Hausdorff, hence normal, we find open $V_c \subset C$ such that $c \in V_c \subset \overline {V_c} \subset U_c$ and a continuous Urysohn-function $u_c : C \to [0,1]$ such that $u_c(c) = 1$ and $u_c(x) = 0$ for $x \in C \setminus V_c$. Define $$g_c : C \to \mathbb R^{k(c)} \times \mathbb R = \mathbb R^{k(c)+1} , g_c(x) = \begin{cases} (u_c(x) h_c(x), u_c(x) )& x \in \overline {V_c} \\ (0, u_c(x)) = (0,0) & x \in C \setminus V_c \end{cases}$$ Both parts of the definition yield continuous functions on the closed subsets $\overline {V_c}$ and $C \setminus V_c$ of $C$, respectively. On $\overline {V_c} \cap (C \setminus V_c) = \operatorname{bd} V_c$ both parts agree, therefore $g_c$ is continuous.
Let $W_c = u_c^{-1}((0,1])$. This is an open subset of $C$ such that $c \in W_c \subset V_c$.
Claim: If $g_c(x) = g_c(y)$ and $x \in W_c$, then $x = y$.
Proof. $g_c(x) = g_c(y)$ implies $u_x(x) = u_x(y)$. For $x \in W_c$ we have $u_c(x) > 0$, thus also $u_c(y) > 0$ which shows $y \in W_c$. Hence $x,y \in W_c \subset V_c$ and thus $u_c(x) h_c(x) = u_c(y) h_c(y)$. Since $u_c(x) = u_c(y) > 0$, we conclude that $h_c(x) = h_c(y)$ and therefore $x = y$.
Since $C$ is compact, finitely many $W_{c_i}$ cover $C$. Define $$h : C \to \prod_{i=1}^n \mathbb R^{k(c_i)+1} = \mathbb R^N, h(x) = (g_{c_1}(x), \ldots, g_{c_n}(x)) .$$
We shall check that $h$ is injective; thus it is an embedding because $C$ is compact. So let $x, y \in C$ such that $h(x) = h(y)$. We have $x \in W_{c_i}$ for some $i$, thus $g_{c_i}(x) = g_{c_i}(y)$ with $x \in W_{c_i}$. Our above claim shows that $x = y$.