# Connected subsets

﻿Keywords: connected subsets
Description: Use that $X$ is connected $\iff$ the only continuous functions $f:X\to\$ are constant, where $\$ is endowed with the discrete topology. Now, you know each $F$ in $\mathscr F$ is connected.

Use that $X$ is connected $\iff$ the only continuous functions $f:X\to\<0,1\>$ are constant, where $\<0,1\>$ is endowed with the discrete topology.

Now, you know each $F$ in $\mathscr F$ is connected. Consider $f:\bigcup \mathscr F\to\<0,1\>$, $f$ continuous.

Take $\alpha \in\bigcap\mathscr F$. Look at $f(\alpha)$, and at $f\mid_:\bigcup \mathscr F\to\<0,1\>$ for any $F\in\mathscr F$.

Since you mention metric spaces, I am not sure if you know about the first thing I mention, so let's prove it:

THM Let $(X,\mathscr T)$ be a metric (topological) space. Then $X$ is connected if and only if whenever $f:X\to\<0,1\>$ is continuous, it is constant. The space $\<0,1\>$ is endowed with the discrete metric (topology), that is, the open sets are $\varnothing,\<0\>,\<1\>,\<0,1\>$.

The statement is of the form $P\iff (A\implies B)$. First, we prove $\neg P\implies A\wedge \neg B$, then we prove $A\wedge \neg B\implies \neg P$. Note we use $(A\implies B)\equiv \neg A\vee B$, and the usual De Morgan Laws.

P First, suppose $X$ is disconnected, say by $A,B$, so $A\cup B=X$ and $A\cap B=\varnothing$, $A,B$ open. Define $f:X\to\<0,1\>$ by $$f(x)=\begin1& \; ; x\in A\\0&\; ; x\in B\end$$

Then $f$ is continuous because $f^<-1>(G)$ is open for any open $G$ in $\<0,1\>$ (this is simply a case by case verification), yet it is not constant. Now suppose $f:X\to\<0,1\>$ is continuous but not constant. Set $A=\=f^<-1>(\<1\>)$ and $B=\=f^<-1>(\<0\>)$. By hypothesis, $A,B\neq \varnothing$. Morover, both are open, since they are the preimage of open sets under a continuous map, and $A\cup B=X$ and $A\cap B=\varnothing$. Thus $X$ is disconnected. $\blacktriangle$