Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this 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
add comment

2 Answers

up vote 3 down vote accepted

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 $A$.

share|improve this answer
add comment

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 $A$?

share|improve this answer
    
I see the tag edit has happened. I'll remove these no longer relevant comments. –  user21436 Apr 20 '12 at 3:40
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.