# $\mathbb{N}$, with an open set defined as a set with a finite complement, is not path connected. [duplicate]

I want to prove that the topological space $\mathbb{N}$, with an open set defined as a set with a finite complement, is not path connected.

In case it is relevant, I have already shown (at least I hope so) that this space is connected and compact.

My plan is to suppose that we have a continuous function $$f: [0,1] \to \mathbb{N}$$ such that $f(0)=n$ and $f(1)=m$, where $m,n\in\mathbb{N}$, and then produce a contradiction.

It seems to me that the definition of continuity I need is probably the one that says that the open sets in the range must come from open sets in the domain, or else the one that says that closed sets in the range must come from closed sets in the domain.

Other than that, I am a bit stuck.

Any tips would be greatly appreciated.

*BEGIN EDIT: Thanks to David Moews comments, I can see now how the problem reduces to showing that $[0,1]$ cannot be written as the union of countably many disjoint closed sets. On the MathOverflow post referred to in the comments, I found this sketch of a proof:

"The proof I like uses the fact that a nested sequence of open intervals has non-empty intersection provided neither end point is eventually constant. Now one inductively constructs a sequence of such intervals as follows. Each interval is a component of the complement of the union of the first n closed sets, for some n. Then wait till the next closed set intersects that interval. (If it never does, then we're trivially done.) It cannot fill the whole interval, and indeed must miss out an interval at the left and an interval at the right. So pass to one of those subintervals in such a way that your left-right choices alternate. Done."

I am not sure if I understand this argument correctly. Here is my version of it:

We know that for $n=f(0)$ and $m=f(1)$ there are closed sets $[0,a]$ and $[b,1]$ in $[0,1]$, possibly singleton, such that $f([0,a])=n$ and $f([b,1])=m$. The union of these two closed subsets $[0,a]\cup [b,1]$ is a closed subset of $[0,1]$, with the open subset $(a,b)$ as its complement. Suppose that $[0,1]$ can be written as a union of countably many closed and disjoint sets. Then there is a closed set $[c,d]$ that lies within $(a,b)$. Since there exists $x$ such that $a=Inf((a,b))<x<c$, etc, we now have $[0,1]=[0,a]\cup (a,c)\cup [c,d]\cup(d,b)\cup [b,1]$. Similarly, there must be a closed set $[e,f]$ that lies within $(a,c)$ so that $(a,c)=(a,e)\cup [e,f]\cup (f,c)$. Note that $(f,c)\subset (a,b)$. We can proceed inductively so that $(f_{n+1},c_{n+1})\subset (f_{n},c_{n})\subset ...\subset (a,b)$. So there is always another open set, and if we can show that this set contains a point, then there is a point that is not in the union of countable closed and disjoint sets and we have a contradiction.

I'm not quite sure how to show this but perhaps I could argue:

Since $(a,...,f_n,f_{n+1},...)$ is strictly increasing and $(b,...,c_n, c_{n+1},...)$ is strictly decreasing and since $\forall n, f_n\leq c_n$, they are bounded monotone sequences of real numbers and must converge. Thus we have our point.

Is this line of thinking correct. Or have I misunderstood something?

I'm sorry if I am going over old territory, but I just want to make sure I am on the right track. Thanks in advance for any help. *END EDIT

## marked as duplicate by YuiTo Cheng, Yanior Weg, José Carlos Santos general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 8 at 20:43

• So, filling in the dots, it seems that $\forall n\in\mathbb{N}, \{n\}$ is a closed set. So that $f^{-1}(n)$ must be closed in [0,1], by assumption of continuity. Then since there are countably many elements in the range of $f$, [0,1] must be the union of countably many closed sets. These sets are non empty, since for each $n\in\mathbb{N}$ there is at least one $x\in [0,1]$ such that $f(x)=n$, and disjoint, since $f$ cannot send the same $x_{0}$ to two distinct elements of $\mathbb{N}$. – jim Mar 16 '13 at 3:20
• Yes, that's right. A priori, you could also have a partition of $[0,1]$ into only a finite number ($\ge 2$) of nonempty closed sets, but this would contradict the fact that $[0,1]$ is connected. – David Moews Mar 16 '13 at 3:35