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Let $X$ be a topological space. I wonder what are the possible ways frequently used by people to prove the little property: Let $B$ be closed in $X$, $A\subseteq B$. Then $\overline{A}\subseteq B$.

My accustomed way is: Let $x\in\overline{A}$. Then $x$ is an adherent point of $A$ (i.e. every open neighborhood of $x$ intersect $A$). Since $A\subseteq B$, $x$ is clearly and trivially an adherent point of $B$. Since $B$ is closed, it contains all its adherent point. Therefore $x\in B$. However, I feel that many people tend to think about "closed" as "its complement is open", "closure" as "the intersection of all closed sets containing that set". In metric space, even some sequentially characterization of "closed set" is preferred. So I want to ask how would you do to prove such little theorem. I want to compare and learn from it. (I'd tried to prove using another way, but the proof get quite lengthier.)

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    $\begingroup$ I'd go with the intersection of closed supersets definition $\endgroup$ – Hagen von Eitzen Jun 30 at 2:50
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    $\begingroup$ The sequential definition only works in nice spaces, like first-countable ones, not in all spaces. So most texts avoid it, unless the goal is to do analysis only. It can be saved by going to nets, or filters, but that's often a distraction at an early stage where texts want to discuss closures. $\endgroup$ – Henno Brandsma Jun 30 at 8:33
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The proof you gave via adherent points is fine.

I like as the definition of closure:

$$\overline{A}=\bigcap \{F: A \subseteq F \text{ closed in } X\}$$

because it's then more obvious that the closure is the smallest closed set that contains $A$, closer to the intuition you expressed.

If $A \subseteq B$ and $B$ is closed, the asked for inclusion is immediate: we can take $F=B$ as one of the sets we take the intersection of (!) and as the intersection is a subset of any of its intersecting sets by definition, $\overline{A} \subseteq B$ is immediate.

Or an alternative using adherent points: suppose $x \notin B$, then $B^\complement$ is an open set (as $B$ is closed) that misses $A$, so $x$ is not an adherence point of $A$, hence $x \notin \overline{A}$. This shows the inclusion by contrapositive.

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  • $\begingroup$ Thanks! Pretty clear. $\endgroup$ – Eric Jun 30 at 12:18
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Using any definition you can easily see two things:

a) $A \subset B$ implies $\overline A \subset \overline B$

b) $\overline B=B$ if $B$ is closed.

Putting these two together we get the result.

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