11 - Connectedness of Intervals

Connectedness of Intervals

Any interval $[x,y]$ in $\mathbb{R}$ is connected

Note that IVP is a consequence of this theorem and Connected image of a connected set

Proof

Using Connectedness via open separations
Suppose $[x,y]\subseteq U\cup V$ where $U$ and $V$ are open subsets in $\mathbb{R}$ with $[x,y]\cap U\cap V=\varnothing$
And $[x,y]⊈U$ and $[x,y]⊈V$ thus

$$[x,y]\cap U\text{ and }[x,y]\cap V\text{ are both non-empty}$$

Without loss of generality assume $x\in [x,y]\cap X$ and $y\in [x,y]\cap V$

Why $y\in [x,y]\cap V$

As $[x,y]\cap V\ne \varnothing$ there exists $y^{\prime }\in [x,y]$ with $y^{\prime }\in V$
Note $y^{\prime }\ne x\in U$

Replacing $[x,y]$ by possibly smaller interval $[x,y^{\prime }]$ then

$$[x,y^{\prime }]\subseteq U\cup V\text{ with }[x,y^{\prime }]\cap U\cap V=\varnothing$$

Moreover

$$[x,y^{\prime }]⊈U\text{ and }[x,y^{\prime }]⊈V\text{ as }x\in U\text{ and }{y}^{\prime }\in V$$

This leads to a contradiction to show assumption hence $y\in [x,y]\cap V$

Note that as $[x,y]\cap U\cap V=\varnothing$ then $y\notin U$ and $x\notin V$

Define $S=\{z\in [x,y]:z\in U\}$ then $S$ is non-empty and bounded so

$$c=\sup (S)\text{ exists}$$

Note that $c\in [x,y]$ and since $[x,y]\subseteq U\cup V$ then either $c\in U$ or $c\in V$

Assume $c\in U$ so as $y\notin U$ then $c\ne y$
As $U$ is open then there is some interval $[c,c+ϵ)$ contained in $U$ and also in $[x,y]$
However $[c,c+ϵ)\subseteq S$ but contradicts that $c=\sup (S)$

Similarly assume if $c\in V$ then $c\ne x$
As $V$ is open there is some interval $(c-ϵ,c]$ contained in $V$ and in $[x,y]$
In particular as $\left[c-\frac{ϵ}{2},c\right]$ is disjoint from $S$ this contradicts $c=\sup S$

Hence as we have the two contradictions then neither $[x,y]\subseteq U$ or $[x,y]\subseteq V$ hence

$$[x,y]\text{ is connected}$$