2

Let $(X,d)$ be a metric space and $A$ is a subset of $X$.

$A^c$ is complement of $A$ in $X$.

Use only the following characterization of closed sets: $$A \text { is closed if it contains all it's limit points}$$ and show that $A^c$ closed $\Rightarrow$ for all $a\in A$ there exists $r>0$ such that $B(a,r)$ is contained in A.

My attempt: Suppose $a\in A$. Then $a\notin A^c$.

So $a$ is not the limit of any sequence in $A^c$. So we have that there exists $\epsilon>0$ such that $d(a,x)>\epsilon$ for all $x\in A^c$ so we can choose $r = \epsilon/2$.

Is this correct?

fosho
  • 6,334

1 Answers1

1

Let it be that $a\in A$ and that no $r>0$ exists with $B(a,r)\subseteq A$.

Then for every $r>0$ the set $B(a,r)\setminus\{a\}$ contains an element of $A^c$.

Then by definition $a$ is a limitpoint of $A^c$.

It is not an element of $A^c$ so we conclude that $A^c$ is not closed.

drhab
  • 151,093
  • Thanks for this. However is there a problem with my approach? – fosho Mar 22 '16 at 09:14
  • @Hetebrij no it does not since no matter what $x$ you choose in $A^c$, we have $d(a,x)>\epsilon$. – fosho Mar 22 '16 at 09:16
  • @Daniel I suggest a rewording: "we have that there exists $\epsilon>0$ such that for all $x$..", which solves this problem and makes sense. Your original statement here, if not wrong, really is misleading. – Vim Mar 22 '16 at 09:17
  • @Vim, yes this is a great suggestion. – fosho Mar 22 '16 at 09:17
  • Now your approach seems okay to me. At first hand @Hetebrij was correct, though. That's also why I gave this alternative. – drhab Mar 22 '16 at 09:23
  • @drhab, thanks, I much prefer your solution! – fosho Mar 22 '16 at 09:23