# 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_

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)=\begin

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=\