# Basic Topology: Connectedness

Suppose that E is a connected proper subset of R^n and that E is a proper subset of A which is a proper subset of the closure of E. Prove that A is connected.

I can't seem to really find a starting point. My general approach to showing a set is connected is to either reach a contradiction assuming it is disconnected or to show that the set is convex (and thus connected).

Hints are greatly appreciated!

Our definition of connectedness is: A set E is connected if and only if it cannot be separated by any pair of relatively open sets. Otherwise, it is disconnected.

The word 'proper' can be omitted at both places and the result is true in any topological space. Here is the proof: suppose $A=C\cup D$ where $C$ and $D$ are disjoint open sets in $A$. Then $E=(E\cap C) \cup (E\cap D)$. From the definition of subspace topology it is clear that $E\cap C$ and $E\cap D$ are open in $E$. By connectedness of $E$ we get $E\cap C$ or $E\cap D$ is empty. Suppose $E\cap C$ is empty. We claim that $C$ is empty in this case. Write $C=C\cap U$ where $U$ is open in the whole space. If possibel, let $x \in C$. Then $x$ is in the closure of $E$ and $U$ is an open set containing $x$. Hence there is a point $y$ in $E\cap U$. But then $y$ is in $E \cap C$ which is a contradiction. Thus $C$ is empty in this case. Similarly, in the remaining case $D$ is empty. We have proved that $A$ is connected.
Let $B_1$ and $B_2$ be two open balls whose adherence are disjoint. $x_1\in B_1$, $E=B_1-\{x_1\}\cup B_2$, $A=B_1\cup B_2$, $E$ is a proper subset of $A$ and $A$ is a proper subset of the adherence of $E$ which is the disjoint union of the adherence of $B_1$ and $B_2$, but $A$ is not connected.
Suppose $f \colon A\to\{0,1\}$ is a continuous map.
The restriction of $f$ to $E$ is constant, without loss of generality we can assume $f(x)=0$ for $x\in E$.
Can it be $f(a)=1$, for some $a\in A$?