See if function is homeomorphism - "lege artis" answer 
$X=\mathbb{N}_0$ and  $Y= \left\lbrace \frac{1}{n} ~|~ n\in \mathbb{N}  \right\rbrace \cup \left\lbrace 0\right\rbrace $ are subspaces of euclidean space $\mathbb{R}$. See if function $f: X \to Y $ 
$  f(n) =
  \begin{cases}
     \frac{1}{n} &  n\in \mathbb{N}\\
    0  &   n =0 \\
  \end{cases}
$
is a homeomorphism.

How would you do this textbook style, leaving little unsaid and being formal as possible by using only beginner theorems? I know how to do this but my proof always involve some handwaving. All my topology is handwaving, to be honest all my proofs are handwaving, I never mastered the art of a good explanation. For once I just want to see an obnoxious proof. Thanks  
 A: There cannot be an homeomorphism between these two spaces, because $Y$ has a limit point in $Y$ (this limit point is $0$) while $X$ has no limit point in $X$. 
A: Let $f$ be a bijection between $X$ and $Y$. For simplicity, name $g=f^{-1}$. 
Consider $a=g(0)=f^{-1}(0) \in \mathbb{N}$, and consider the set $\{a\}$ which is equal to $\mathbb{N} \cap (a-(1/2),a+(1/2))$.
$(a-1/2),a+(1/2))$ is an open interval, and those form a base of open sets in standard topology on $R$, so clearly the interval itself is an open set. By definition, intersection of an open interval and a subset is a open set in the subspace topology of that specific subset - so we get that $\{a\}$ is an open set in the subspace $\mathbb{N}$ of $\mathbb{R}$. 
If $g$ is to be continuous, from the definition of continuity, we have that for each $x \in Y$, and each open neightborhood $G$ of $g(x)$, there must exist an open neightborhood $F$ of $x$, such that the image of this neightborhood with g must be a subset of $G$. So for $x=0$  there must exist an open set $U$ in the space $Y$ containing $0$, such that the image of this set with $g$ is a subset of $\{a\}$, as we have shown that is an open neightborhood of $a$. 
But every open set contaning $0$ also contains another point:
Take an open set $A$ containing $0$, and because open intervals form a base of the standard topology on $\mathbb{R}$, we can find an open interval $(x,y)$ inside $A$ containing $0$ - clearly $y > 0$, and because for each real number we can find a larger integer, take $n> (y^{-1})$ - we have now found an $n \in \mathbb{N}$ such that $n^{-1} < y$, so every open set in $\mathbb{R}$ also contains a point $n^{-1} $ for some $n \in \mathbb{N}$, distinct from $0$, and because this point is in $Y$, we have also shown we can find such a point, distinct from $0$, in any open set of $Y$ subspace of $\mathbb{R}$ containing $0$.
Because $g$ is a bijection, the image of any open neightborhood of $0$ in $Y$ will contain at least two elements, and so it cannot possibly be a subset of $\{a\}$, therefore $g$ is not continuous, and so $f$ is not a homeomorphism, as $f^{-1} = g$.
Thus, no bijection between $X$ and $Y$ is a homeomorphism, and so a homeomorphism between $X$ and $Y$ does not exist. In particular, the function 
$f(n) =
  \begin{cases}
     \frac{1}{n} &  n\in \mathbb{N}\\
    0  &   n =0 \\
  \end{cases}$
is not a homeomorphism.
A: Assume $f^{-1}: Y\to X$ is continuous. Clearly, it maps $0\in Y$ to $0\in X$. One definition of continuity in topology is that preimages of open sets are open.
Open sets in $X$ resp. $Y$ are such sets that with each $x$ also contain all points closer to $x$ then $\epsilon$ for some $\epsilon>0$. (This is co-called metric topology induced by the standard distance function on $\Bbb R$. It's the most obvious topology on $X,Y\subseteq\Bbb R$ one can think of, unless otherwise stated.)
If you accept this, and accept that the one-point set $\{0\}$ is open in $X=\Bbb N_0$, then $\{0\}$ should be open in $Y$. But it is not, because any neighborhood of $0$ in $Y$ contains infinitely many other points.
Of course, if you want to be very formal, one should probably say, in the assignment, "$X,Y$ are topological subspaces of $\Bbb R$ with the metric/Euclidean topology". If you endow $\Bbb R$ with a discrete topology---that is, all subsets are open---then $f$ is a homeomorphism, because it is a bijection and preimages of "open" (that means, any) sets under $f$, resp. $f^{-1}$ are "open" (that means, any) sets.
A: If you're endowing these with the subset topology, then every point of $X$ is open, since each point is the intersection of some open set in $\mathbb R$ and all of $X$.  For $Y$ every point is an open set, except $0$.  You'd expect continuity problems with this.  For example $f(\{0\})$ maps an open set to a non-open set.  So $f^{-1}$ can't be continuous, and $f$ can't be a homeomorphism.  (All homeomorphisms are open maps.)
A: *

*Any subset of a metric space is also a metric space under the induced metric.

*If $f^{-1}$ was continuous, then $0=f^{-1}(0)=f^{-1}(\lim 1/n)=\lim f^{-1}(1/n)=\lim n$.
This is a contradiction.
