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If the closure of a set $A$ is defined as the intersection of all closed sets which contain $A$, prove that closure of a closed set $B$ is $B$ itself.

Attempt: I apologize if this is too basic but I am taking an inttoductory course in Elements of Real Analysis.

Let $B$ be a closed set, then closure of a $B$ is defined as the intersection of all closed sets which contain $B$.

Hence, we wants closed sets $B_1,B_2, \cdots , B_m$ such that $B_1 \bigcap B_2 \bigcap \cdots \bigcap B_m = B$

Such that $B \subseteq B_1,B \subseteq B_2,~~\cdots~~, B \subseteq B_n$

We need to prove using the given definition that $B_1 \bigcap B_2 \bigcap \cdots \bigcap B_m = B$

Since, intersection of closed sets is a closed set $\implies B_1 \bigcap B_2 \bigcap \cdots \bigcap B_m = A$ where $A$ is a closed set and $B \subseteq A$ ( Where $B$ is also a closed set )

How do I move ahead using the given definition. I am aware of the method of solving the problem using the derived set method, but the problem seeks to solve it through the given definition only.

EDIT : Since, one of the closed sets that contains B is B itself, then is it appropriate to take $B_1=B_2=⋯B_m=B $. But, then what about the other closed sets which contain $B$?

Thank you for your help.

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  • $\begingroup$ Note that in general there will be infinitely many closed sets that contain $B$, not just $m$ of them. $\endgroup$ – Greg Martin Sep 2 '14 at 7:58
  • $\begingroup$ Uhm, yeah .. But, since intersection of any number of closed sets is a closed set, will it make a difference? $\endgroup$ – MathMan Sep 2 '14 at 8:00
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    $\begingroup$ It doesn't make sense to take $B_1 = B_2 = \ldots = B_m = B$. The question asks you to show something about the intersection of every closed set containing $B$. One of these will be $B$ itself, and then there will usually be some other closed sets containing $B$ that are not $B$ itself. The claim is that $B$ is equal to the intersection of all of these. $\endgroup$ – MJD Sep 2 '14 at 14:39
  • $\begingroup$ Oh okay I got it.. Thank you :-) $\endgroup$ – MathMan Sep 2 '14 at 14:42
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This is merely a question of logic. It has not much to do with the exact definition of closedness, nor with finite vs. infinite intersections, etc.

If $B$ is a closed set and $B$ its closure according to definition then $$B\subset\bar B\subset B\ ,\tag{1}$$ and this implies $B=\bar B$. In $(1)$ the first inclusion holds because all sets $A$ entering the intersection $\bar B$ contain $B$, and the second inclusion holds because $B$ itself is one of these sets $A$.

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  • $\begingroup$ I got it. Thank you for your answer :-) $\endgroup$ – MathMan Sep 2 '14 at 15:06
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Hint: one of the closed sets that contains $B$ is $B$ itself.

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  • $\begingroup$ I was actually a little confused regarding this. Since, one of the closed sets that contains $B$ is $B$ itself, then is it appropriate to take $B_1=B_2= \cdots B_m = B$ . But, then what about the other closed sets which contain $B$? $\endgroup$ – MathMan Sep 2 '14 at 8:02
  • $\begingroup$ Am I thinking on the correct lines? Thanks $\endgroup$ – MathMan Sep 2 '14 at 8:12
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Here's a second hint: Suppose you have two sets with different definitions, as you do here. Let's call them $X$ and $Y$. Then to show that $X=Y$, it is enough to show $X\subseteq Y$ and $Y\subseteq X$. To show $X\subseteq Y$ you take an arbitrary point $p$ in $X$ and show that $p$ is also a point of $Y$; to show $Y\subseteq X$ you do the reverse.

In this case you want to show $$\bigcap_{\substack{B\subseteq C\\\text{$C$ closed}}} C = B.$$ Let's call the thing on the left $X$ for short. You then want to show $X=B$.

To do this, it is enough to show that $X\subseteq B$ and that $B\subseteq X$. Start with $X \subseteq B$. Suppose some point $p$ is in $X$. What properties must $p$ have? Can you use those properties to show that $p$ must also be in $B$?

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  • $\begingroup$ Yes, I got it. Thank you for your answer :-) $\endgroup$ – MathMan Sep 2 '14 at 15:05

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