[Math] What does this mean? $ A \subset \bar{A} \subset F $ where $F$ is a closed set containing $A$.

general-topologyreal-analysis

I'm having a little trouble visualizing this relation. But here's my thought process:

  1. $\bar A$ is the closure of A; i.e. the intersection of all closed sets in A. It is also the smallest closed superset of A.
  2. F is a closed set containing A.

When I see such a relation I figure the following to some extent must be true:

$ A \le \bar{A} \le F$: I understand how $F$ could be larger than $A$ but I don't think I understand why the closure of $A$ ($\bar A $) could be larger than $A$? The closure should be the intersection of all the closed sets in $A$ so why would such an intersection be larger than set $A$? That's the bit I can't quite grasp.

Is there some sort of diagram or more clear proof that I should be looking at or…?

Best Answer

The obvious example would be $\overline{(0,1)}=[0,1]=(0,1)\cup\{0,1\}$, as Daniel Fischer notes in the comment.

But note that $\overline{\Bbb Q}=\Bbb R$, so the closure can not only be strictly larger as a set, but can be much larger in cardinality as well.

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