# Punctured plane is homeomorphic to the plane without the unit disk.

I'm trying to prove that the function $$f(r,\theta) = (r+1,\theta)$$ is a homeomorphism between the punctured plane and the plane minus the closed unit disk using polar coordinates.

I got a proof to show that $$f^{-1}$$ is continuous but it doesn't seem correct and I'm not sure what the issue is. It goes like this:

Let $$A$$ be an open set in the punctured plane and let $$f(r,\theta)\in f(A)$$. We want to find a $$z$$ small enough such that $$B(f(r,\theta),z)\subset f(A)$$ to show that $$f(A)$$ is open. Note that $$(r,\theta) \in A$$ thus $$B((r,\theta),\epsilon) \subset A$$ for some $$\epsilon$$.

Now consider the set $$X=\{d(f(r,\theta),f(p,\phi))\in \mathbb{R}:(p,\phi) \notin B((r,\theta),\epsilon)\}$$. Clearly $$X$$ is a nonempty subset of real numbers and is bounded below by $$0$$. Thus we can set $$z = \inf X$$. Consider $$B(f(r,\theta),z)$$. If $$f(a,\psi) \in B(f(r,\theta),z)$$ then $$d(f(a,\psi),f(r,\theta)) which implies $$d(f(a,\psi),f(r,\theta)) \notin X$$ and thus $$(a,\psi) \in B((r,\theta),\epsilon)$$. So $$(a,\psi) \in A$$ and $$f(a,\psi) \in f(A)$$ so $$B(f(r,\theta),z) \subset f(A)$$. Therefore $$f(A)$$ is open and $$f^{-1}$$ is continuous.

The issue I have with this proof is that I made no mention to the actual definition of $$f$$ however I cannot find the flaw or step where a mistake was made for this proof the be invalid.

• actually with the plane minus the closed unit disc Commented Aug 9 at 17:24
• I suspect that with a bit of thought you could write a formula for $f^{-1}$, which would aid your proof of continuity. Commented Aug 9 at 19:10

Your worries are well-founded: the proof is incorrect. To see why, forget about the function $$f$$ for a moment and consider an arbitrary subset $$U\subset\mathbb R^2$$. Your argument essentially goes as follows:

Let $$x\in U$$, and choose $$\epsilon=\operatorname{inf}\{d(x,y):y\notin U\}.$$ Then $$B(x,\epsilon)\subseteq U$$, so $$x$$ is in the interior of $$U$$.

But of course this "proves" that every set is open, which we know is false. So what's the problem? I recommend choosing $$U$$ to be a set you know isn't open, and carefully following the argument to see where it fails.

Hint:

For simplicity's sake, take $$U=\{x\}$$ to be a single point. What does $$\epsilon$$ equal in that case?

Full explanation:

Even though $$\{d(x,y):y\notin U\}$$ consists only of numbers $$>0$$, its infimum can still equal 0. In fact, we will have $$\epsilon=0$$ if and only if $$x$$ is not in the interior of $$U$$. In order to use this argument to show that a set $$U$$ is open, we would have to somehow prove that $$\epsilon>0$$.

Well, you have to use the definition of $$f$$ to at least witness $$f$$ is bijective. Then in your proof, you make the following key oversight: you must check $$z>0$$ ! But you did not. Really, this is the whole game for showing $$f$$ is an open map / $$f^{-1}$$ is continuous. You can use the definition for that... You also should say a few words. Even granted that $$z>0$$, your proof doesn't immediately show $$f(A)$$ contains $$B(f(r,\theta),z)$$. You merely showed $$f(A)$$ contains $$B(f(r,\theta),z)\cap\mathrm{Im}(f)$$. Maybe you noticed this, in which case I apologise, but it's a meaningful point; $$f$$'s surjectivity is needed for your argument - you only test set membership against elements of the form $$f(a,\psi)$$, rather than generic elements $$(b,\vartheta)$$.

The main thing for you to do now is prove $$z>0$$.

• @JacksonSmith No worries. I should have said in my post - the reason you need $z>0$ is so that $B(f(r,\theta),z)$ is actually open Commented Aug 9 at 17:59