Basic Topology: Connectedness

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $a\in A$?