Vrftsjtryk

Consider the set $N=mathbb Ncup{infty}$ together with the following topology: a subset $U$ of $N$ is open if either $inftynotin U$ or $Nsetminus U$ is finite.

I'm trying to show that $A={frac{1}{n}:nge 1}cup{0}$ with topology induced from $mathbb R$ is homeomorphic to $N$.

Define the map $f:Nto A, nmapsto 1/n$ for $nin mathbb N={1,2,dots}$ and $infty mapsto 0$. It is bijective because its inverse is given by $g: Ato N, 0mapsto infty, 1/nmapsto n (nge 1)$. So it suffices to show that $f$ and $g$ are continuous.

Continuity of $f$. It suffices to show the preimage of any basic open set is open. Any basic open set is of the form $Acap U$ where $Usubset mathbb R$ is open. If $U$ does not contain $0$, then its preimage doesn't contain $infty$, so it's open. Now assume $U$ contains $0$. We need to show its preimage has finite complement. How to prove this? My argument shows that this is not true: $N-f^{-1}(Acap U)=f^{-1}(mathbb R-(Acap U))$, and $mathbb R-(Acap U)$ is infinite (since $Acap U$ is at most countable).

Continuity of $g$. Let $Usubset N$ be open. First suppose $infty notin U$. Then $0notin g^{-1}(U)$, but I'm stuck here. The second case is when $N-U$ is finite. This doesn't tell me anything about openness of the preimage either...

First note that $N$ is compact: let $mathcal{U}$ be an open cover of $N$. Let $U_0in mathcal{U}$ be an open set that contains $infty$. This means that $Nsetminus U_0$ is finite (as it cannot be of the first type of open set that does not contain $infty$). For every of the finitely many $n in N setminus U_0$ we pick some $U(n)in mathcal{U}$ that contains $n$ and then $U_0, {U(n): n in Nsetminus U_0}$ is the required finite subcover of $mathcal{U}$.

So as the codomain is Hausdorff (even metric), it suffices to show continuity of $f$ to see it is a homeomorphism, as it will be a closed map by compactness.

Let $O$ be open in ${frac{1}{n}: n ge 1} cup {0}$.

If $0 in O$ then as in the reals $frac{1}{n} to 0$, we know that for some $M in mathbb{N}$ we have that $n ge M$ implies $frac{1}{n} in O$, by the definition of convergence. So $f^{-1}[O]$ contains ${n: n ge M} cup {infty}$ which ensures that $N setminus f^{-1}[O] subseteq {n : n < M}$ is finite and so $f^{-1}[O]$ is open.

If $ 0 notin O$ then $infty notin f^{-1}[O]$ by definition, and so $f^{-1}[O]$ is also open. So in either case, $f$ is continuous. Note that we could have replaced ${frac{1}{n}: n ge 1} cup {0}$ by any infinite convergent sequence in the reals and its limit in the place of $0$. The checking of continuity is the same.