Given $\mathbb{N}_{CF}$, let $A\subset\mathbb{N}$ be a finite set such that $A\neq \emptyset$. Prove: $\mathbb{N}_{CF}\diagup A \cong \mathbb{N}_{CF}$

Question

Given $\mathbb{N}_{CF}$, the cofinite topology on $\mathbb{N}$. let $A\subset\mathbb{N}$ be a finite set such that $A\neq \emptyset$.

Prove: $\mathbb{N}_{CF}\diagup A \cong \mathbb{N}_{CF}$ hence, the quotient space $\mathbb{N}_{CF}\diagup A$ is homeomorphic to $\mathbb{N}_{CF}$

I'm not sure I understand what needs to be proved. if $A$ is finite, then $|\mathbb{N}\setminus A| = |\mathbb{N}$, as $A\neq \emptyset$,and $A$ is finite, why is the argument not immediate by cardinality considerations?

Answer 1

$$|\Bbb N\setminus A|=|\Bbb N|$$

Isn't the whole story. For two spaces to be homeomorphic, it is necessary but not at all sufficient that the underlying sets have the same cardinality. For example, if you give the set $\{0,1\}$ the discrete topology, the trivial topology and the Sierpinski space topology, you have three spaces of equal cardinality and all of them are distinct: no one is homeomorphic to the other. A more interesting example is that the unit circle $S^1$ is not homeomorphic to $[0,1)$ despite there being a continuous bijection $[0,1)\to S^1$.

The quotient space is defined here how it is always defined:

$\Bbb N/A$ has point set the set $(\Bbb N\setminus A)\sqcup\{A\}$ where $\{A\}$ is just a canonical way to adjoint a singleton. Ordinarily I'd just write $\ast$ for some arbitrary single point, which is to represent $A$. We write $q:\Bbb N\to\Bbb N/A$ as the map $n\mapsto n$ - if $n\notin A$ - and $n\mapsto\{A\}$ - if $n\in A$.

The topology on $\Bbb N/A$ is precisely the set of all $U$ for which $q^{-1}(U)$ is open in $\Bbb N$.

That definition is formally precise but also a little unwieldy. Just think of $\Bbb N/A$ as the space you get when you take $\Bbb N$ and 'continuously' squash $A$ down to a single point, but leave everything else the same. Neighbourhoods of this new single point must un-squash to neighbourhoods of the set $A$.

Hopefully that clears things up.

As for the exercise, here is a hint:

The map $q$ has a certain special property, that whenever $X$ is any other space and $f:\Bbb N\to X$ is continuous, if $f$ is constant on $A$ then $f=gq$ for a unique continuous $g:\Bbb N/A\to X$. In particular if you want to find a homeomorphism with $\Bbb N$, it's actually sufficient to find a map $\Phi:\Bbb N\to\Bbb N$ with this same property (why? Think about taking $X:=\Bbb N$ and $X:=\Bbb N/A$).

Think about $\Phi$ which involve bijections of sets $\Bbb N\setminus A\cong\Bbb N$. Does any such $\Phi$ have this property?