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• 1 Post By mfb

# Thread:

1. ## Is this set open?

A is a subset of B, both sets are open.
Is the set B/A open?

My answer would be no, because the complement of an open set is always closed. Is this the correct definition in this case?

2. ## Re: Is this set open?

But a set can be both open and closed.

You can look at A=(0,1), B=(0,2) as subset of the real line with the usual topology.

3. ## Re: Is this set open?

girdav: In your example, A\B is [1,2) which is not open.

However, let B be the union of two arbitrary disjoint open subsets A and C. B is open.
Then B\A = C is open by construction.

4. ## Re: Is this set open?

But why is the set open? If it's the complement of A and A is open, shouldn't the complement of an open set always be closed?

5. ## Re: Is this set open?

Originally Posted by infernalmich
But why is the set open? If it's the complement of A and A is open, shouldn't the complement of an open set always be closed?
You must understand that $B\setminus A=B\cap A^c$ and is known as a relative complement.
That is the complement of A relative to B.
You see $A$ is an open set in the underlying space which may not be $B$.
In the real numbers $((0,2)\setminus (0,1)=[1,2)$ which is not open.

6. ## Re: Is this set open?

Originally Posted by infernalmich
But why is the set open? If it's the complement of A and A is open, shouldn't the complement of an open set always be closed?
Yes, that's true. But "B\A" is NOT the "complement" of A, which is, by definition, the set all points not in A, whether they are in B or not. If $B= (0, 1)\cup (2, 3)$ and $A= (0, 1)$, then $B\A= (2, 3)$ an open set. The [b]complement of A is $(-\infty, 0]\cup [1, \infty$, a closed set.
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