$\overline{X\cap Y}\subset \overline{X}\cap\overline{Y}$ for real numbers, case when $\overline{X\cap Y}\neq \overline{X}\cap\overline{Y}$

Question

My proof for this is similar to this one, but I can't find an example such that $\overline{X\cap Y}\neq \overline{X}\cap\overline{Y}$ for the real numbers.

Is there in any way implied that $X \neq Y$? Otherwise I don't see how you could prove it or if it is even true. It could be that I have the wrong definition of closure though (if it is closure you are implying in the question, the bar could also mean complement) — coolcat007, May 30 '16 at 04:55
Nvm, I found a good example here: https://math.stackexchange.com/questions/77373/give-an-example-that-overlinea-cap-b-neq-overlinea-cap-overlineb — coolcat007, May 30 '16 at 05:07

score 5 · Answer 1 · answered May 30 '16 at 04:10

5

The rationals and the irrationals will do.

answered May 30 '16 at 04:10

Henno Brandsma

242,131

score 3 · Accepted Answer · answered May 30 '16 at 04:14

3

Easy example: $X = (0,1), Y = (1,2)$. Their intersection is empty, but the intersection of their closures is not.

answered May 30 '16 at 04:14

Eric Towers

67,037

$\overline{X\cap Y}\subset \overline{X}\cap\overline{Y}$ for real numbers, case when $\overline{X\cap Y}\neq \overline{X}\cap\overline{Y}$

2 Answers2