Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 19 July 2014

Lecture 31

Let $X=ℝ\text{.}$ Let $E\subseteq ℝ\text{.}$ Then $E$ is connected if and only if $E$ is an interval.

		Proof.
	To show: $E$ is not connected if and only if $E$ is not an interval. To show: (a) If $E$ is not connected then $E$ is not an interval. (b) If $E$ is not an interval then $E$ is not connected. (a) Assume $E$ is not connected. Then there exist open sets $U_1,U_2$ such that $$U_1\cup U_2\supseteq E,U_1\cap E\ne \varnothing ,U_2\cap E\ne \varnothing \text{and}(U_1\cap E)\cap (U_2\cap E)=\varnothing \text{.}$$ Let $x\in U_1\cap E$ and $y\in U_2\cap E$ and suppose $x<y\text{.}$ Let $$z=\text{sup}(U_1\cap E\cap [x,y])\text{.}$$ Then $$x\le z<y\dots$$ (b) To show: If $E$ is not an interval then $E$ is not connected. Assume $E$ is not an interval. Then there exist $x,y\in E$ with $x<y$ and $z\notin E$ with $x<z<y\text{.}$ Let $U_1=(-\infty ,z)$ and $U_2=(z,\infty )\text{.}$ Then $U_1,U_2$ are open, $U_1\cup U_2\supseteq E,$ $x\in U_1\cap E,$ $y\in U_2\cap E$ and $(U_1\cap E)\cap (U_2\cap E)=\varnothing \text{.}$ So $E$ is not connected. $\square$

A Hausdorff space is a topological space $X$ such that if $x,y\in X$ and $x\ne y$ then there exists a neighbourhood $N_x$ of $x$ and a neighbourhood $N_y$ of $y$ such that $N_x\cap N_y=\varnothing \text{.}$

The notes for Lectures 29 and 30 have a slight error in the definition of not connected. The correct definition is:

Let $X$ be a topological space.
Let $E\subseteq X\text{.}$
The set $E$ is not connected if there exist open sets $U_1$ and $U_2$ such that $$U_1\cup U_2\supseteq E,U_1\cap E\ne \varnothing ,U_2\cap E\ne \varnothing ,\text{and}(U_1\cap E)\cap (U_2\cap E)=\varnothing \text{.}$$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100521Lect31.pdf and was given on 21 May 2010.

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