Prove that if all set homeomorphic to $X$ is bounded, then $X$ is compact.

Question

Let $X \subset \mathbb{R}^{n}$. Prove that if all set (subsets of $\mathbb{R}^{n}$) homeomorphic to $X$ is bounded, then $X$ is compact.

Let $\varphi: Y \to X$ be an homeomorphism. Since $Y$ is bounded, $Y \subset B(0,\delta)$ for some $\delta >0$. If $X$ is unbounded, for all $\delta_{i} > 0$, there is a $f(y_{i}) \in X$ such that $f(y_{i}) \not\in B(0,\delta_{i})$. Thus, we obtain a sequence $f(y_{i})$ such that $f(y_{i}) \to \infty$. But, $f^{-1}$ is continuous, so $y_{i} \to \infty$. Contradiction. That make sense?

Let $f(y_{i})$ a sequence in $X$ such that $f(y_{i}) \to a$. We have that $f^{-1}$ is a continuous function, so $$\lim f(y_{i}) = a \Longrightarrow \lim f^{-1}(f(y_{i})) = f^{-1}(a),$$ then $y_{i} \to f^{-1}(a)$, but $Y$ is not necessarily compact. How I show this $a \in X$?

Answer 1

Since you have showed that $X$ is bounded, it suffices to argue that $X$ is closed.

Assume that $X$ is not closed. Then there is $z\notin X$ and $x_n \in X$ such that $x_n \longrightarrow z$.

Then we define $ f: X\to\mathbb{R}^n $, given by $f(x)=\frac{1}{z-x}$, notice that $f$ is continuous and injective, so $f$ is invertible. Furthermore, $f^{-1}(y)= z-\frac{1}{y}$ which is also continuous. Therefore, $f(X)$ is homeomorphic to $X$.

But, since there is $x_n\longrightarrow z$, $f(X)$ is unbounded, which leads to the desired contradiction.

Answer 2

I'll prove the equivalent statement that if $X \subset \mathbb{R}^n$ is not compact then it is homeomorphic to an unbounded set.

If $X$ is already unbounded, we're done. So, we can assume that $X$ is bounded. Since $X$ is bounded but not compact, there exists a point $P \in \mathbb{R}^n - X$ such that $P \in \overline X$.

Let $\mathbb{R}^n \cup \{\infty\}$ denote the one point compactification, which is homeomorphic to $S^n$. Let $f : \mathbb{R}^n \cup \{\infty\} \to S^n$ be a homeomorphism. Since $S^n$ is a homogeneous topological space, there exists a homeomorphism $g : S^n \to S^n$ such that $g \circ f(P)=f(\infty)$. It follows that $f^{-1} \circ g \circ f(P)=\infty$. Letting $R = f^{-1} \circ g \circ f(\infty)$ it follows that $G = f^{-1} \circ g \circ f$ is a homeomorphism from $\mathbb{R}^n - \{Q\}$ to $\mathbb{R}^n - \{R\}$. By restriction, $G$ maps $X \subset \mathbb{R}^n - \{Q\}$ homeomorphically to $G(X) \subset \mathbb{R}^n-\{R\}$.

Now we just have to prove that $G(X)$ is unbounded. Since $P \not\in X$ but $P \in \overline X$, we may choose a sequence of points $x_i \in X$ which converges to $P$. It follows that $g \circ f(x_i)$ coverges to $g \circ f(P)$, and so $G(x_i)$ converges to $G(P)=\infty$ in the one point compactification $\mathbb{R}^n \cup \{\infty\}$. The sequence $G(x_i) \in G(X)$ therefore has norms $|G(x_i)|$ approaching $\infty$ and so $G(X)$ is unbounded.

Answer 3

You already know that $X$ is bounded. It remains to show that if $(x_n)$ is a sequence in $X$ which converges to some $x_\infty \in \mathbb{R}^n$, then $x_\infty \in X$ (i.e. that $X$ is complete). Consider the translation $t : \mathbb{R}^n \to \mathbb{R}^n, t(x) = x - x_\infty$; it is a homeomorphism. $(t(x_n))$ is a sequence in $t(X)$ converging to $0$. Assume $x_\infty \notin X$. Then $0 \notin t(X)$. Define a homeomorphism $h : \mathbb{R}^n \backslash \lbrace 0 \rbrace \to \mathbb{R}^n \backslash \lbrace 0 \rbrace, h(x) = x/ \lVert x \rVert^2$ ($h$ is ist own inverse). Then $h(t(X))$ is a homeomorphic copy of $X$ which is unbounded since $h(t(x_n)) \to \infty$, a contradiction.