A question from my analysis homework:
Let $A \subset M$ be connected and contain more than one point. Show that every point in $A$ is an accumulation point of $A$.
This question makes sense to me intuitively because if there existed an $x \in A$ such that $x$ was not an accumulation point of $A$, then we can draw an $\epsilon$-ball around $x$ such that the only element from $A$ the ball contains is $x$ (i.e. $x$ is "isolated"), which means there is empty space between $x$ and the rest of $A$, contradicting the hypothesis. However, I am having trouble finding a proof.
My approach has been this, suppose there exists an $x \in A$ such that $x$ is not an accumulation point of $A$. I want to find two open sets $U$ and $V$ such that
- $U \cap V \cap A = \emptyset$
- $U \cap A \ne \emptyset$
- $V \cap A \ne \emptyset$
- $A \subset U \cup V$
The intuitively obvious choice for these open sets is $U = D(x,\epsilon)$, $V = A \setminus \{x\}$, where $\epsilon$ is the radius for which $A \cap D(x,\epsilon)$ = {x}. However, it may not be the case that $A \setminus \{x\}$ is open, which is the only condition preventing this proof from working. Is there a nice work-around that fixes this issue? Any hints would be appreciated.
$V = M \setminus\{x\} $ is open.