# Union of a connected set and its accumulation point [closed]

Let $A$ be a connected set in the metric space $(X, d)$. If $p$ is an accumulation point of $A$,then prove that $B = A \cup \{p\}$ is connected.

-

## closed as off-topic by Behaviour, Miha Habič, Jyrki Lahtonen, dragon, Sami Ben RomdhaneJul 30 at 7:59

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Behaviour, Miha Habič, Jyrki Lahtonen, dragon, Sami Ben Romdhane
If this question can be reworded to fit the rules in the help center, please edit the question.

homework is not a stand alone tag on this site. Please use another subject tag. BTW, what have you tried? –  user21436 Apr 20 '12 at 3:13
Do you mean to say $B=A\cup {p}$? –  Han Altae-Tran Apr 20 '12 at 3:14
@HanAltae-Tran. How would you define $A\cup p$ since $p$ is not a subset but an element of $X$? –  Thomas E. Apr 20 '12 at 4:22
The question before was written as $B=AU{p}$. Of course, we must only consider unions of sets. –  Han Altae-Tran Apr 20 '12 at 23:24

Suppose the contrary: there exists a separation $B_{1},B_{1}$ of $U:=A\cup \{p\}$. Wlog we may assume that $p\in B_{1}$. If $A\cap B_{1}\neq\emptyset$ then we have a contradiction since $A$ is connected, hence $A\cap B_{1}=\emptyset$. Since $B_{1}$ is open in $U$ there exists $r>0$ so that $B_{U}(p,r)\subset B_{1}$, which is a neighborhood of $p$ that does not intersect $A$. This is a contradiction since $p$ was an accumulation point of $A$.
Suppose you have a separation. Then $p$ is contained in a nontrivial clopen set. What else can you say about this set? What can you conclude about $A$?