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

Question

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 $d(f(a,\psi),f(r,\theta))<z$ 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.

Answer 1

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 $\epsilon>0$.

Answer 2

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 $z>0$.