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Prove: $A\cup A'$ is the smallest closed set containing $A$, i.e.if $F$ is closed and $A\subset F\subset A\cup A'$ then $F = A\cup A'$

Definitions: $A'$ is the set of all accumulation or limit points.

proof: Let $F$ be closed and $A\subset F\subset A\cup A'$. Let $a\in F$. Since $F$ is closed, then $a$ is an limit point for $F$. Thus, by the definition of limit point there exists an open set $G$ containing $a$ contains a different point $q$ of $F$, where $a\neq q$. So we have, $$ q\in G\subset F$$ but, $A\subset F$ and $a$ is an accumulation point of $F$, then $F = A\cup A'$

I am not sure if I am right, any suggestions would be greatly appreciated

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  • $\begingroup$ Why is it not converting into latex? $\endgroup$ – Wolfy Mar 24 '15 at 13:20
  • $\begingroup$ It's not true that a point $a\in F$ is a limit point. $\endgroup$ – egreg Mar 24 '15 at 13:21
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It's not true that if $a\in F$ then it is a limit point.

You're starting from the wrong side: you have to show that points in $A\cup A'$ are also in $F$.

If $a\in A\cup A'$, then either $a\in A$ or $a\in A'$.

In the first case there's nothing to prove. In the second case, $a$ is a limit point of $A$, hence also a limit point of $F$. Since $F$ is closed, you have $a\in F$.

Note that if $A\subset B$, then all limit points of $A$ are necessarily limit points of $B$.

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First just a couple points about your proof:

Just because $a\in F$ doesn't necessarily make it a limit point of $A$. So then your conclusion that $F=A\cup A'$ is not correct.

My suggestion is to start by letting $a$ be a limit point of $A$. Then you can conclude things about $A'$ by whether or not $a$ is in $A$ or $A'$.

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