3
$\begingroup$

I proved $A \subseteq \overline{A}$ by showing $\overline{A}^c \subseteq A^c$: Let $x \in \overline{A}^c $ then $x$ is not an accumulation point of $A$ therefore there is an open nbhood $N$ of $x$ with $N \cap A = \varnothing$ or equivalently: $N \subseteq A^c$.

This is fairly short. I tried do it directly but failed. Is it possible to show $A \subseteq \overline{A}$ directly? (Assume $x \in A$ and $N$ an open nbhood of $x$. Then $N$ must contains a point of $A$)

The definition of closure is the set and all its accumulation points.

$\endgroup$
  • 5
    $\begingroup$ How do you define the closure of a set? Depending on it, the answer might be more straightforward than another. $\endgroup$ – Pedro Tamaroff Feb 28 '13 at 17:18
  • $\begingroup$ @PeterTamaroff I added definition to question! $\endgroup$ – blue Feb 28 '13 at 17:24
5
$\begingroup$

If you've defined $\overline A$ to be the set $A$ together with all its accumulation points, then by definition, $A$ is a subset of $\overline A.$

If instead you've defined it to be the set of all $x$ in the overlying space such that every neighborhood of $x$ intersects $A,$ then note that for any $x\in A$ and any neighborhood $N$ of $x$ we have $x\in N,$ so $A\cap N\neq\emptyset,$ and so $x\in \overline A.$

It's a good exercise to show that these two definitions are equivalent. (Hint: $x$ is an accumulation point of $A$ if and only if every neighborhood of $x$ intersects $A$ at some point different from $x$.)

$\endgroup$
  • $\begingroup$ Why do I not have to show that $a \in A$ is an accumulation point of $A$? $\endgroup$ – blue Feb 28 '13 at 17:28
  • 2
    $\begingroup$ @blue Not really. $a\in A$ might be isolated, but it will still be part of the closure of $A$. $\endgroup$ – Pedro Tamaroff Feb 28 '13 at 17:30
  • $\begingroup$ @PeterTamaroff You both are right, I am feeling stupid. The definition says the closure is all accumulation points together with all the points of the set. Thank you people! $\endgroup$ – blue Feb 28 '13 at 17:31
  • $\begingroup$ @blue No reason to feel stupid! Cheers. $\endgroup$ – Pedro Tamaroff Feb 28 '13 at 17:32
4
$\begingroup$

It depends on what specific definition you are taking for $\overline A$. For example, the definition I would take is that $\overline A$ is the intersection of all the closed sets that contain $A$. Then as each of the sets you are intersecting contains $A$ we can conclude that their intersection $\overline A$ also contains $A$.

$\endgroup$
3
$\begingroup$

Let $A'$ denote the set of accumulation points of $A$. Then you've defined $$\bar A=A\cup A'$$

This means $\bar A\supseteq A$ at once.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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