TITLE: Prove a square is homeomorphic to a circle
QUESTION [22 upvotes]: $s:=\{|x|\le 1,|y|\le 1\} $
$c:=\{{x}^{2}+{y}^{2}\le1\}$
Prove $\overset{\circ}{s}  \cong \overset{\circ}{c}$

ok... not to sure what to do.
I think $\overset{\circ}{s}  \to\overset{\circ}{c}$ is something like:
$$(x,y)\rightarrow\left(\frac{x}{\sqrt{{x}^{2}+{y}^{2}}},\frac{y}{\sqrt{{x}^{2}+{y}^{2}}}\right)$$
What is the inverse for this? Do i need an inverse? Do i just prove the function is continous and the inverse is continuous?
Please help...

REPLY [2 votes]: You can, in fact, prove that any bounded convex open subset of $R^n$ is homeomorphic to $R^n$, thus any two bounded convex open subsets of $R^n$ are homeomorphic to each other. The interiors of a circle and a square are bounded convex open subsets of $R^2$, so this general answer also answers the specific question. My proof sketch goes as follows, although it's possible there might be a mistake. Let $A$ be a bounded convex open subset of $R^n$.

WLOG $0 \in A$, since if not, we can just translate.

Let $B(A)$ be the boundary of $A$. For any nonzero $x \in R^n$ define $L(x) := \{cx:c \in R^+\}$ i.e. $L(x)$ is the ray in the direction of $x$. Prove: $B(A) \cap L(x)$ is a singleton for all $x$. As a corollary, if $\alpha \in B(A) \cap L(x)$, then $cx \in A$ if $|cx| < |\alpha|$ and $cx \not\in A$ if $|cx| \geq |\alpha|$.

Define $g:R^n\backslash\{0\} \rightarrow R^n$ by $g(x)$ is the element in $B(A) \cap L(x)$. Prove $g$ is continuous.

Define $h:A \rightarrow R$ by $h(0) = 1$, and for $x$ nonzero, $h(x) = \frac{|g(x)|}{|g(x)|-|x|}$. Then define $f:A \rightarrow R^n$ by $f(x) = h(x)x$. Prove $h,f$ are continuous.

Finally, we prove that $f$ is the desired homeomorphism by finding the inverse function and proving it's continuous. The inverse can be found algebraically. You can use the corollaries from part 2 to prove that the inverse does indeed map into $A$.


Note that this proof sketch assumes a decent undergraduate level understanding of analysis.