The exercise asks to prove that any pair of three sets $$A=\Bbb{N}\\A'=\{0\}\cup\{\frac{1}{n} | n\in \Bbb{N}\}\\ A''=\{0\}\cup\{\frac{1}{n} | n\in \Bbb{N}\} \cup \Bbb{N}$$ is not homeomorphic.
I proved it for $(A,A'), \ (A',A'')$ using compactness of $A'$ and non compactness of the rest, but I struggle with proving the result for the pair $(A, A'')$.
The second part of the question asks to prove $A\times A'$ and $A' \times A''$ are homeomorphic.And with this part I also do have problems.
On
Let's say that $f:A''\to A$ is a homeomorphism. Since $a_n=\frac{1}{n}$ is a convergent sequence in $A''$ then $f(a_n)$ is convergent in $A$. But since $A$ is discrete then convergent sequences in $A$ are precisely sequences which are eventually constant. Meaning that $f(a_n)$ is an eventually constant sequence which is a contradiction because $f$ is injective.
The second part of the question asks to prove $A\times A'$ and $A' \times A''$ are homeomorphic.And with this part I also do have problems.
Right, so this isn't true (thanks @JustinYoung). Every homeomorphism maps non-isolated points to non-isolated points. In particular if $A\times A'$ and $A'\times A''$ are homeomorphic then so are their respective sets of non-isolated points.
But a point in $A\times A'$ is non-isolated if and only if it is of the form $(n,0)$. Therefore the space of all non-isolated points of $A\times A'$ is discrete.
On the other hand the space of all non-isolated points of $A'\times A''$ has at least $(0,0)$ and $(0, \frac{1}{n})$. And thus it is not discrete.
On
I think the following is true:
if X,Y are homeomorphic topological spaces, with homeomorphism h:X$\rightarrow$Y, and if z $\in$ X is a limit point of X, then h(z) is a limit point for Y.
Since 0 is a limit point for A'', but it is not a limit point for A, then A and A'' are not homeomorphic.
It has been several years since I studied math, so excuse me if I am incorrect.
A is discrete, not compact. A' is compact, not discrete.
A" is not compact and not discrete.