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Show that if $U$ is open and $A$ is closed, then $U\setminus A = \{ x\in U : x\notin A \}$ is open. What can be said about $A\setminus U$

I dont quite get why $U\setminus A = \{x\in U : x\notin A\}$ is open?

If $x\in U$ and $x\notin A$ then isn't $U\setminus A$ with just be $U$?

They dont even have common element in the set?

When they dont have common element, isnt $A\setminus U$ will just be the same???


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Tip: rewrite A\U as A\setminus U. – Git Gud Feb 24 '13 at 15:53
up vote 11 down vote accepted

Hint: rewrite $U\setminus A=U\cap A^c$. Now if $A$ is closed, what does that mean for the complement $A^c$?

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I guess $A^c$ is open, and $U$ is open too. Thats why $U\setminus{A}$ is open? – Paul Feb 24 '13 at 16:01
@Paul Yes, by the requirements of a topology, any finite intersection of open sets is open. – 1015 Feb 24 '13 at 16:05

$U \setminus A$ is, in words, all those elements of $U$ that aren't element of $A$. So if $A$ and $U$ were disjoint then this would be equal to $U$, but certainly not in general. If $A = X$ (where $X$ is the whole space) then no elements of $U$ wouldn't be elements of $A$ and the set would be empty. So it does depend on $A$.

Now $U \setminus A = U \cap (X \setminus A)$ (this is immediate from my description in words) and then we have written it as the finite intersection of an open set $U$ and the complement of a closed set $A$, so another open set. So the result must be open.

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A set is closed iff its complement is open.

The intersection of two open sets is open.

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