# The Definition of a closure point.

If I said that $x \in \overline{A}$ (closure of A), does this mean that $\forall G_{x}$ (means an open set containing x) $G_{x}\cap A \neq \varnothing$?

Is not any interior point satisfies this also ?

• Yes, that is correct. – drhab May 1 '17 at 12:40
• But is not any interior point satisfies this also ? – Intuition May 1 '17 at 16:09
• Yes but interior points also satisfy stronger conditions the boundary points fail to satisfy. – avs May 1 '17 at 16:12
• Read Kolmogorov and Fomin's "introduction to real analysis". – avs May 1 '17 at 16:13
• Yes, they do, the set of interior points of $A$ is a subset of $A$ which on its turn is a subset of $\overline{A}$. – drhab May 1 '17 at 16:15

This definition of $\overline{A}$ is quite correct, it's the set of adherence points of $A$ i.e. $$\overline{A} = \{x \in X: \forall O \subseteq X \text{ open } : x \in O \rightarrow O \cap A \neq \emptyset \}$$
If $x \in A'$, so $x$ is an accumulation point of $A$, i.e $$A'= \{x \in X : \forall O \subseteq X \text{ open }: x \in O : O \cap (A \setminus \{x\}) = (O \setminus \{x\} ) \cap A \neq \emptyset\}$$
It's clear that $A' \subseteq \overline{A}$: if $O \setminus \{x\}$ intersects $A$, then also the larger set $O$ will intersect $A$.
$A$ is closed iff $A = \overline{A}$ iff $A' \subseteq A$. Both are often used as the definition of a closed set. Also, if $x \in \overline{A}$ and $x \notin A$ then $x \in A'$, also clear from the definitions.
It's also clear that $\operatorname{int}(A) \subseteq A \subseteq \overline{A}$ I don't see any contradiction there.
• about your first definition of $\bar{A}$: what is $X$ and what is $O$? Is $O$ an open ball centered at $x$? – johnny09 Apr 20 at 17:49
• @johnny09 $X$ is the whole space we’re working in. $O$ is any open set containing $x$; it could be a an open ball centered om $x$. – Henno Brandsma Apr 20 at 18:28