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I have to prove:

$(X\cap Y)^- \subset X^- \cap Y^-$

Well, if $a\in (X\cap Y)^-$ then there is an open set $A$ containing $a$ such that:

$$A\cap (X\cap Y)\neq \emptyset$$

I've tought of some distritution property but it would work if I had to prove an equality, but here I need to prove an inclusion.

So, if $a\in (X\cap Y)^-$, then $d(a, X\cap Y)=0\implies$ there is $z\in X\cap Y$ such that $d(a,z)=0$. But $z\in X$ and $d(a,z)=0$, and , $z\in Y$ and $d(a,z)=0$, therefore we should have $z\in X^-\cap Y^-$, right? Any way to explain the end in a more logical way?

Guerlando OCs
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  • Is "$A^-$" the closure of $A$? – MPW Apr 13 '16 at 03:39
  • If it is, then you may want to use \overline{X \cap Y} to get "$\overline{X \cap Y}$". – user21820 Apr 13 '16 at 03:42
  • Whenever $X\cap Y \ne \phi$ there is always an open set $A$ with $a\in A $ such that $A\cap (X\cap Y)\ne \phi$ because we may take $A$ to be the whole space...... A point $p$ is in the closure of $S$ ifff EVERY open set containing $p$ intersects $S$.... You can also use " \bar x " for a single character (" x ") to write $\bar x $. – DanielWainfleet Apr 13 '16 at 03:53
  • See also: http://math.stackexchange.com/questions/599460/prove-that-the-closure-of-the-intersection-of-a-and-b-is-the-subset-of-the-inter – Martin Sleziak Apr 13 '16 at 05:20

1 Answers1

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You are assuming the wrong thing. If $a \in \overline{X \cap Y}$, it means that every open set around $a$ intersects $X \cap Y$. To see this definition, look at this page : http://mathworld.wolfram.com/SetClosure.html , the fourth definition, and the word neighbourhood can be replaced with open set.

$x \in \overline{X \cap Y}$, then for all open sets $A$ containing $x$, $A \cap X \cap Y \neq \phi$. Of course, this also means that $A \cap X \neq \phi$,whence $x \in \overline{X}$ ,and $A \cap Y \neq \phi$,whence $x \in \overline{Y}$, hence $x \in \overline{X} \cap \overline{Y}$. The converse is not true, as can be seen with $X=(0,1)$ and $Y=(1,2)$.

Just to point out, if the definition of closure was : there is some open set A containing $a$ such that $A \cap X \neq \phi$ then imagine what the closure of $(0,1)$ would be. Take some point, say $777777$. Take the open interval $(-1000000,1000000)$, and it is an open set containing $777777$, and intersecting (infact containing) $(0,1)$. So $777777$ would be in the closure of $(0,1)$. That would be absurd. Hence you should be careful with les définitions.

Sarvesh Ravichandran Iyer
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