# Connectedness

• Mar 15th 2010, 05:10 AM
Showcase_22
Connectedness
Quote:

Prove that any infinte set with co-finite topology is connected.
Does this question even make sense?

Let $\displaystyle (X,T)$ be an infinite space with co-finite topology.
Let $\displaystyle (U,V)$ be a two point discrete space where $\displaystyle U=\{0,1\}$.

Let $\displaystyle f: X \rightarrow U$ be any continuous function.

I need to show that f is constant.

My big problem is this:

We know $\displaystyle f^{-1}(1)$ and $\displaystyle f^{-1}(0)$ are open.

Hence $\displaystyle f^{-1}(0), \ f^{-1}(1) \in T$.

However, $\displaystyle X-T$ is finite so there exist finitely many points $\displaystyle x_1, \ldots , x_n$ where$\displaystyle f(x_i) \neq 0$ or $\displaystyle 1$ where $\displaystyle i=1, \ldots , n$.

By that reasoning, we have a finite number of points where the function is not defined. These points are in $\displaystyle X$ so the function is discontinuous at each $\displaystyle x_i$.

Why is this right or wrong?
• Mar 15th 2010, 07:39 AM
Tinyboss
Suppose an infinite set $\displaystyle X$ with cofinite topology were disconnected. Then there are open subsets $\displaystyle U,V$ such that $\displaystyle U\cap V=\varnothing$ and $\displaystyle U\cup V=X$. In particular, we have that both $\displaystyle U$ and $\displaystyle X\setminus U$ are open in the co-finite topology. Is that possible?
• Mar 15th 2010, 07:59 AM
Showcase_22
No! that's not possible! =D

If a set $\displaystyle U$ is open, it's complement should be closed! (That I do know!)

I was misinterpreting what a co-finite topology was. I didn't realise that $\displaystyle X \setminus U$ was also open.
• Mar 15th 2010, 09:06 AM
Drexel28
Quote:

Originally Posted by Showcase_22
No! that's not possible! =D

If a set $\displaystyle U$ is open, it's complement should be closed! (That I do know!)

I was misinterpreting what a co-finite topology was. I didn't realise that $\displaystyle X \setminus U$ was also open.

The cofinite topology is a neat one in the sense that it cannot support two disjoint open sets. Thus it is also, in fact, a $\displaystyle T_1$ space which is not Hausdorff. Also, it is clear that since every totally disconnected space is Hausdorff that a cofinite space is also not totally disconnected (this was immediately obvious before actually doing the question was my point)

A more interesting question you may want to ask yourself: is a cofinite space every locally connected?