# Direct proof of $A \subseteq \overline{A}$

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.

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

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 $x$.)

• Why do I not have to show that $a \in A$ is an accumulation point of $A$? – blue Feb 28 '13 at 17:28
• @blue Not really. $a\in A$ might be isolated, but it will still be part of the closure of $A$. – Pedro Tamaroff Feb 28 '13 at 17:30
• @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! – blue Feb 28 '13 at 17:31
• @blue No reason to feel stupid! Cheers. – Pedro Tamaroff Feb 28 '13 at 17:32

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 $A$.

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.