A compact, Hausdorff and connected topology on $\mathbb {N}$ I am studying Topology for the first time and I am attempting to solve the following homework problem.

Given an example of a topology on $\mathbb {N}$ such that it makes it a compact and Hausdorff space. Can this topology also be connected?

My solution so far: Let $(\mathbb{N},\tau)$ and $(\mathbb{N}$$\cup\left\{ \infty\right\} ,\tau')$
be two topological spaces. The topology $\tau$ is the subspace topology
inherited from the usual topology in $\mathbb{R}$ and $\tau'$ is
the topology of the Aleksandrov compactification of $\mathbb{N}$.
Consider the map 
\begin{array}{cccc}
f: & (\mathbb{N}\cup\left\{ \infty\right\} ,\tau^{'}) & \rightarrow & (\mathbb{N},\tau^{*})\\
 & \infty & \mapsto & 0\\
 & n & \mapsto & n+1
\end{array}
I claim that the final topology induced by $f$ is Hausdorff and compact. 
$\tau$ is the discrete topology since every singleton is open.
$\tau'=\tau\cup\{(\mathbb{N}\setminus C)\cup\{\infty\}\mid C\text{ is compact in }\mathbb{N}\}$.
The compact sets of $\mathbb{N}$ are finite sets because the compacts
sets of a discrete topology must be finite (the proof being clear).
Therefore, 
$$
\tau'=\tau\cup\left\{ (\mathbb{N}\setminus\{n,\dots,k\})\cup\{\infty\}\right\} 
$$
The final topology induced by $f$ is the finest topology that makes
$f$ continuous, i.e. $U\in\tau^{*}\iff f^{-1}(U)\in\tau'$ and it
follows that 
$$
\tau^{*}=\left\{ \left\{ n\right\} \mid n\in\mathbb{N}\setminus\left\{ 0\right\} \right\} \cup\left\{ \mathbb{N}\setminus\left\{ n,\dots,k\right\} \mid n>0\right\} 
$$
It's easy to see that $(\mathbb{N},\tau^{*})$ is compact since for
any cover of $\mathbb{N}$ there is a finite subcover of the form
$\mathbb{N}\setminus\left\{ n,\dots,k\right\} \cup\left\{ n,\dots,k\right\} $.
I could also reason that the image of a compact space is compact but
one gains more insight in this fashion. 
Let's check that $(\mathbb{N},\tau^{*})$ is still Hausdorff. Let
$x,y\in\mathbb{N}$ such that $x\neq y$. There are two cases to be
considered. 


*

*If $x,y\neq0$ then pick the following open sets $U=\left\{ x\right\} $
and $V=\left\{ y\right\} $. 

*Now suppose, without loss of generality that $x\neq0$. Then choose
the opens sets $U=\left\{ x\right\} $ and $V=\mathbb{N}\setminus\{x\}$.
In either case $U\cap V=\emptyset$ and $U,V\in\tau^{*}$proving that
it is indeed Hausdorff.


Clearly this topological space isn't connected. Given any subspace
$U\subset\mathbb{N}$ which is not a singleton or $\emptyset$ (these
are always trivially connected) it is always possible to find $V_{1},V_{2}\in\tau^{*}$
such that $V_{1}\cap V_{2}=\emptyset$ and $V_{1}\cup V_{2}=U$. Again,
let's consider two different cases. 


*

*If $0\notin U$ then $U$ is a discrete space which is trivially disconnected. 

*If $0\in U$ then simply set $V_{1}=U\setminus\left\{ 0\right\} $
and $V_{2}=\mathbb{N}\setminus\left\{ n,\dots,k\right\} $ such that
$\left\{ n,\dots,k\right\} \notin V_{1}$. 


This proves that this space is actually totally disconnected. 
Questions: I feel that this is always true, i.e. given any compact and Hausdorff topology on $\mathbb{N}$ it is always the case that it can't be connected. My argument would go like this: since the Aleksandrov compactification is unique up to homeomorphism then every topology that I could give for $\mathbb{N}$ would give rise to a homeomorphic topological space. Clearly I could have defined $f$ in another way but the topology would be the "same" (up to homeomorphism). I found this exercise in a book and the hint they gave was: think of Baire category theorem. But I honestly can't see the connection. I know that a compact and Hausdorff space is a Baire space but what follows from here...?  
 A: Depending on your background, you may find the following approach to be easier than using Baire's theorem.

A nontrivial connected $T_{3.5}$ space (where $T_{3.5}$ means completely regular and $T_1$) is uncountable.

Proof:
Let $X$ be a nontrivial connected $T_{3.5}$ space. Since $X$ is nontrivial it contains two distinct elements $x$ and $y$.
Due to the fact that $X$ is $T_1$, the singleton $\{y\}$ is closed. Then complete regularity implies the existence of a continuous function $f:X\to[0,1]$ such that $f(x)=0$ and $f(y)=1$. The connectedness of $X$ and continuity of $f$ implies that $f(X)$ is a connected subspace of $[0,1]$ containing $0$ and $1$. Therefore $f(X)=[0,1]$, so $X$ is uncountable. $\square$
Now suppose $X$ is a countably infinite compact Haudorff space.
Since compact Haudorff spaces are normal and normal spaces are (by Urysohn's lemma) completely regular, we infer that $X$ is a nontrivial countable $T_{3.5}$ space. The above result applies to conclude that $X$ is not connected.
A: Baire's theorem/the definition of Baire spaces can be stated in terms of intersections of open sets or of unions of closed sets. Here, it's in my opinion more suitable to use the "unions of closed sets" phrasing. A topological space $X$ is a Baire space if and only if for every countable family $\{ F_n : n \in \mathbb{N}\}$ of closed sets with empty interior, the union
$$\bigcup_{n\in \mathbb{N}} F_n$$
has empty interior.
In particular, if we write $X$ - which is assumed to be a Baire space - as a countable union of closed sets,
$$X = \bigcup_{n \in \mathbb{N}} A_n,$$
then at least one of the $A_n$ has nonempty interior.
Now if $X$ is a countable Baire space, we can write it as a countable union of singletons. If $X$ is a $T_1$ space, then singletons are closed, so then it follows that at least one of the singletons has nonempty interior. But the only nonempty subset of a singleton is the singleton itself, so a singleton with nonempty interior must be an open set. And hence, if $X$ contains more than one point, that gives a partition of $X$ into two disjoint nonempty open sets, the open singleton $\{x\}$ and $X\setminus \{x\}$, so $X$ is not connected.
A (locally) compact Hausdorff space is a Baire space, and clearly $T_1$, so a connected compact Hausdorff space with more than one point must be uncountable.
A: Your example is fine, I would just pick a countable compact Hausdorff space, like the convergent sequence $X=\{\frac{1}{n}: n =1,2,3,4,\ldots\} \cup \{0\}$ and take any bijection with $\mathbb{N}$ and define the topology on $\mathbb{N}$ to be such that $f$ is a homeomorphism, say if $f: X \to \mathbb{N}$ is a bijection, define $O \subseteq \mathbb{N}$ open iff $f^{-1}[O]$ open in $X$.
Note that $X$ is just homeomorphic to the ALexandroff compactification of $\mathbb{N}$, which is your example.
It's also homeomorphic to the ordinal $\omega+1$ in the order topology, and in fact it's not too hard to show that any countable compact Hausdorff space is homeomorphic to some countable ordinal in the order topology. 
Such a space can never be connected because a connected normal ($T_4$) space has size at least continuum. This is because if $x \neq y$ in a normal space we can find a continuous function $f:X \to [0,1]$ such that $f(x) =0$ and $f(x)= 1$,by Urysohn's lemma. By connectedness of $X$, $f[X] \subseteq [0,1]$ is connected and contains $0$ and $1$, so $f[X] = [0,1]$. So $X$ has size at least continuum.
