Difference of closures is subset of closure of difference

[Rasalhague]

I can't find the source of this statement now, but I've been trying to prove that

$$\overline{A}\setminus\overline{B} \subseteq \overline{A\setminus B}.$$

Now $x\in\overline{A}\setminus\overline{B}$ means x is in every closed superset of A but there exists a closed superset of B that doesn't contain x, whereas $x\in \overline{A\setminus B}$ means x is in every closed set that contains every point of A that's not also in B.

I've tried various manipulations involving equivalent definitions of closure, but have yet to find any obvious way to proceed. Any hints?

 

[dextercioby]

The particular case $A\cap B = \varnothing$ is trivial to prove. If x in A but not in B, then it's in A bar and might be in B bar. Let's assume it's not in B bar too, because the LHS would be then the empty set. So if x is in A bar and not in B bar, then x may be in A or not. If it's in A and not in B bar, then x is not in B, too, so that x would be in A and not in B, hence it would automatically be in (A minus B) bar. The only issue is that x is not in A, but is in A bar. So this x cannot be in B bar, hence cannot be in B, thus cannot be in A minus B but it should be in the closure of A minus B, since the closure of A minus B contains the elements of A bar which are not in B (as was our x).

 

[Rasalhague]

Okay, thanks Dexter! Special case: If $A\cap B = \varnothing$, then $A\setminus B = A$ so $\overline{A\setminus B} = \overline{A}$, and $\overline{A}\setminus\overline{B}\subseteq \overline{A}$, so $\overline{A}\setminus\overline{B}\subseteq \overline{A\setminus B}$.

General case: $A = (A\setminus B)\cup (A\cap B)$ so $\overline{A} = (\overline{A\setminus B})\cup (\overline{A\cap B})$.

$$\overline{A}\setminus\overline{B}$$

$$=((\overline{A\setminus B})\cup (\overline{A\cap B}))\setminus ((\overline{B\setminus A})\cup (\overline{A\cap B}))$$

$$=(((\overline{A\setminus B})\cup (\overline{A\cap B}))\setminus (\overline{B\setminus A}))\cap (((\overline{A\setminus B})\cup (\overline{A\cap B})) \setminus(\overline{A\cap B}))$$

$$=(\overline{A}\setminus (\overline{B\setminus A}))\cap(\overline{A\setminus B})$$

$$\subseteq \overline{A\setminus B}. \enspace \blacksquare$$

 

[Rasalhague]

Let's assume it's not in B bar too, because the LHS would be then the empty set.

I didn't follow your meaning here.

 

[dextercioby]

The x was chosen to be in A bar but not in B so it could have been in B bar thus the LHS would have been the empty set, due to the arbitrariness of x.