Can someone help me with proving the following simple relationship:

$\displaystyle

\bar A \setminus \bar B \subset \overline {{A \setminus B}}

$

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- Jan 7th 2008, 10:59 AMDMTProof of closure relationship
Can someone help me with proving the following simple relationship:

$\displaystyle

\bar A \setminus \bar B \subset \overline {{A \setminus B}}

$ - Jan 7th 2008, 11:30 AMPlato
If $\displaystyle x \in \left( {\overline A \backslash \overline B } \right)$ then $\displaystyle \left( {\exists O_x } \right)\left[ {x \in O_x \,\& \,O_x \cap B = \emptyset } \right]$ for some open set.

But any open set that contains*x*must also contain a point of*A*because*x*is in the closure of*A*.

Now if*Q*is any open set containing*x*then $\displaystyle Q \cap O_x $ is also open and as such must contain a point of $\displaystyle A\backslash B$.

Can you finish now. - Jan 7th 2008, 12:05 PMDMT
I think I solved it, but I did it differently, and I don't follow some of your explanation. But maybe I'm confused about something ... do elements of a set have to be either limit points or interior points? This seems to be implied in your explanation, but I thought this wasn't true? (for example the set of natural numbers in the usual topology of Reals).

As I understand it the closure is the union of a set and it's derived set:

$\displaystyle

\bar A = A \cup A^\prime

$

So that your second statement about any open set containing x also containing another point of A doesn't make sense to me ... unless you didn't mean "another" point, but included the possibility of x itself?

I solved it like this. First I changed it to the equivalent:

$\displaystyle

(A \cup A^\prime) \setminus (B \cup B^\prime) \subset (A \setminus B) \cup (A \setminus B)^\prime

$

$\displaystyle

(A \setminus B)^\prime = A^\prime \setminus B^\prime

$

If $\displaystyle

p \in (A \cup A^\prime) \setminus (B \cup B^\prime)

$ then ...

I just realized an error with what I was doing ... I wanted to write next:

$\displaystyle p \in A \setminus B $ or $\displaystyle A^\prime \setminus B^\prime$

But I see now I can't directly rule out $\displaystyle p \in A \setminus B^\prime $ or $\displaystyle A^\prime \setminus B$

And I'm concerned that my derived set equality above is not accurate, though it seemed straightforward a moment ago. I can at least show:

$\displaystyle

(A \setminus B)^\prime \subset A^\prime \setminus B^\prime

$

Though that goes the wrong direction, so I would need equality.

Well, now I've managed to thoroughly confuse myself, so I think I better take a closer look at your method. - Jan 8th 2008, 12:37 AMDMT
I think I got this one now. Either x is in A or A´, if it is in A, then you get directly A\B, otherwise you get (A\B)´ (since there has to be another point of A\B in the open set) ... but it seems to be the introduction of Q was not necessary ... I don't see what this adds. The first open set seems to work fine by itself.