$s:=\{|x|\le 1,|y|\le 1\} $


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...

  • 4
    $\begingroup$ $x^2+y^2 \color{blue}{=} 1$ is a circle. $\endgroup$
    – user17762
    Apr 18, 2013 at 22:09
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    $\begingroup$ Map the boundary of the square to a circle in the obvious way. Then extend this to the interior in the obvious way. $\endgroup$
    – copper.hat
    Apr 18, 2013 at 22:11
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    $\begingroup$ $\overset{\circ}{A}$ is completely standard notation for the interior, used e.g. by Bourbaki. $\endgroup$
    – Martin
    Apr 18, 2013 at 22:15
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    $\begingroup$ @Martin: I’m a retired topologist, and I first saw it here, within the last year. It’s guessable, since $s^\circ$ is common enough, and I gather that it is quite common in some places, but it’s hardly surprising if a U.S. mathematician hasn’t seen it before. $\endgroup$ Apr 19, 2013 at 1:09
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    $\begingroup$ @BrianM.Scott: Then I suppose it is European bias :-) The standard textbooks on topology in French and German do use this notation (the English translation of Jänich's Topology uses it, too). You can find it on the French Wikipedia page, while the German page uses the variant you mention. I always figured the latter notation was due to the difficulties in typesetting the circle in accented position alluded to by kahen. $\endgroup$
    – Martin
    Apr 19, 2013 at 7:01

5 Answers 5


I think the following more general result is true and the argument is also quite simple:

Let $V$ be a real normed vector space and $||\cdot ||_1$, $||\cdot ||_2$ two equivalent norms on $V$, i.e. there are positive constants $C_1, C_2$ such that $||x||_1\le C_1 ||x||_2$ and $||x||_2\le C_2 ||x||_1$. Note that each norm gives rise to the same topology in $V$ because they are equivalent. Define the balls unit balls with respect to each norm as $B_1=\{x\in V:||x||_1\le 1\}$ and $B_2=\{x\in V:||x||_2\le 1\}$. Then $B_1\approx B_2$.

Define $\phi:D_1\to D_2$ as $\phi(x)=\frac{||x||_1}{||x||_2} x$ if $x\neq 0$ and $\phi(0)=0$. It's clear that $\phi$ is continuous in $V-0$ because the norms and the scalar product are continuous functions and continuity at $0$ is also easy to verify using the fact that $||\cdot ||_1$ and $||\cdot ||_2$ are equivalent norms. The inverse is given by $\psi:D_2\to D_1,\psi(x)=\frac{||x||_2}{||x||_1} x$ if $x\neq 0$ and $\psi(0)=0$. So $\phi$ is the desired homeomorphism.

The problem here is a special case of this taking $V=R^2$, the norms $||\cdot||_2$ and $||\cdot||_\infty$ (remembering that all norms in $R^2$, a finite dimensional vector space, are equivalent) and noting that my homeomorphism $\phi$ maps the boundary of the circle to the boundary of the square and so you get an homeomorphism from the interiors by restricting the domain.

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    $\begingroup$ This is an excellent answer for anyone who prefers to rely less on geometric and more on purely analytical arguments. Also quite elegant. No need for complicated formulas. Thanks $\endgroup$ May 26, 2016 at 20:18
  • $\begingroup$ why a homeo of the boundaries would imply homeo of interiors? $\endgroup$
    – Potitov06
    Apr 23, 2021 at 23:13
  • $\begingroup$ @Potitov06 homeo of the boundaries implies that one interior is mapped onto the other interior. Therefore one gets a homeo of interiors by restricting the homeo $\phi$ to the interior of $D_1$. $\endgroup$
    – Zero
    Jun 9, 2021 at 22:30

Note that if we draw any ray outward from the origin, then for any $r\ge 0$ it will intersect precisely one point of the set $$C_r:=\left\{(x,y)\in\Bbb R^2:\sqrt{x^2+y^2}=r\right\}$$ and exactly one point of the set $$S_r:=\bigl\{(x,y)\in\Bbb R^2:\max(|x|,|y|)=r\bigr\}.$$ Every point of each $C_r$ and every point of each $S_r$ is hit by some such ray, and the only point that is hit by more than one ray is the origin, which is the single point of $S_0=C_0$.

Furthermore, $\overset{\circ}{s}$ is the disjoint union of $S_r$ for $0\le r<1$ and $\overset{\circ}{c}$ is the disjoint union of $C_r$ for $0\le r<1$.

A natural candidate for a homeomorphism is to take any point of $\overset{\circ}{s}$, find a ray from the origin that it lies on and the appropriate $S_r$ containing the point, and map the point to the point of $C_r$ on the ray. This is readily a well-defined bijection by the discussion above, as is its inverse (which is basically the same, but maps $C_r$ points to their corresponding $S_r$ points). You need only show that they are both continuous.

It would be simpler to show continuity if we had formulas, though. We already know that the origin will be mapped to the origin. Take $(x,y)\in\overset{\circ}s$ with $(x,y)\ne (0,0)$. Putting $r=\max(|x|,|y|),$ we have $(x,y)\in S_r$. We must map $(x,y)$ to the point $(x',y')$ such that $\sqrt{(x')^2+(y')^2}=r$ lying (not necessarily strictly) between $(x,y)$ and $(0,0)$. That is, we will have $(x',y')=t(x,y)$ for some $0<t\le 1$. In particular, since we need $$r=\sqrt{(tx)^2+(ty)^2}=t\sqrt{x^2+y^2},$$ then we need $$t=\frac{r}{\sqrt{x^2+y^2}}=\frac{\max(|x|,|y|)}{\sqrt{x^2+y^2}},$$ so the map $\overset{\circ}s\to\overset{\circ}c$ is given by $$(x,y)\mapsto\frac{\max(|x|,|y|)}{\sqrt{x^2+y^2}}(x,y)=\frac{|x|+|y|+\bigl||x|-|y|\bigr|}{2\sqrt{x^2+y^2}}(x,y)$$ for $(x,y)\ne0$ and $(0,0)\mapsto(0,0)$. Now, in each variable, this map is continuous for $(x,y)\ne (0,0)$ (as a quotient of a continuous function over a positive continuous function), and so the map is continuous on $\overset\circ s\smallsetminus\{(0,0)\}$. We can show without too much difficulty that $$\frac{|x|+|y|+\bigl||x|-|y|\bigr|}{2\sqrt{x^2+y^2}}(x,y)\to(0,0)$$ as $(x,y)\to(0,0)$, so the map is continuous there, too.

Now take $(x,y)\in\overset{\circ}c$ with $(x,y)\ne (0,0)$. Put $r=\sqrt{x^2+y^2}$. We must map $(x,y)$ to the point $(x',y')$ such that $\max(|x'|,|y'|)=r$ and such that $(x,y)$ lies (not necessarily strictly) between $(x',y')$ and $(0,0)$. Then we need $(x',y')=t(x,y)$ for some $t\ge1,$ so $$r = \max(|x'|,|y'|)= \max(|tx|,|ty|)= \max(t|x|,t|y|)= t\cdot\max(|x|,|y|),$$ so $$t=\frac{r}{\max(|x|,|y|)}=\frac{\sqrt{x^2+y^2}}{\max(|x|,|y|)},$$ and so the map $\overset{\circ}c\to\overset{\circ}s$ is given by $$(x,y)\mapsto\frac{\sqrt{x^2+y^2}}{\max(|x|,|y|)}(x,y)$$ for $(x,y)\ne(0,0)$ and $(0,0)\mapsto(0,0)$. Once again, the map is continuous away from the origin, and using the fact that $$\frac{\sqrt{x^2+y^2}}{\max(|x|,|y|)}\le\sqrt2$$ for all $(x,y)\ne (0,0)$ we can see that the map is continuous at the origin, too.


Hint: The isomorphism is given by $$\phi\colon\overset{\circ}{c}\to\overset{\circ}{s}\colon (x,y)\mapsto\begin{cases} (0,0) & x=y=0 \\\left(\frac{x}{|x|}\sqrt{x^2+y^2},\frac{y}{|x|}\sqrt{x^2+y^2}\right) & |x|\geq|y| \\ \left(\frac{x}{|y|}\sqrt{x^2+y^2},\frac{y}{|y|}\sqrt{x^2+y^2}\right) & |x|<|y|\end{cases}$$

  • $\begingroup$ You're right. I'll fix it. $\endgroup$
    – Abel
    Apr 18, 2013 at 23:59

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 $R^n$.

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

  2. 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 $|cx| \geq |\alpha|$.

  3. 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.

  4. 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.

  5. 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.


You don't need to find the inverse because these two spaces are compact.

  • 4
    $\begingroup$ First of all, no, they aren't (they are bounded, but not closed). Second of all, $[0,1]$ and $\{0\}$ are both compact subsets of the real line. Does that mean they are homeomorphic? In bijective correcpondence? $\endgroup$ Apr 19, 2013 at 1:04
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    $\begingroup$ Cameron: it is a standard result that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Both sets in the OP are closed, compact, and Hausdorff. So a continuous bijection implies a homeomorphism without having to find an inverse and prove it's continuous. $\endgroup$
    – anonymous
    Apr 19, 2013 at 1:21
  • $\begingroup$ @anonymous: An inverse is needed to show a bijection. These sets are not closed; the OP said that the sets are the interiors of the disk and square. $\endgroup$
    – robjohn
    Apr 19, 2013 at 1:57
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    $\begingroup$ You're correct, he did decide to use the interior. You do not have to establish an inverse directly in order to show that a function is a bijection. Moreover, I think what was meant is (as I said) that you don't have to find the inverse and prove that it's continuous. It follows from the above (or would if we were dealing with the closures of the spaces) $\endgroup$
    – anonymous
    Apr 19, 2013 at 4:01

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