I'm taking a college class on Topology and need more help understanding derived points and closure - any help?

Question

So I have to know how to prove whether or not something is a derived point and the definition states that p(exists)X is a derived point of A iff (if and only if) for all epsilon > 0 we have B(p,epsilon) (intersection) A\{p} does not = 0.

Where I find a problem is understanding closure.

The collection of all derived points of A is denoted by A'.
The closure of A is A union A'. A is closed iff A = CL(A).

In my readings I kept finding that a set could be open, closed, both, or neither -- how can this happen and what does this have to do with derived points (and/or also interior points)??

Answer 1

The best way to see if a set is closed is to see if the complement is open.



For example, the complement of [0,1] is (-infty,0)U(1,infty), which is open.



It is a nice exercise to show that this criterion is true. You should prove it.



There are many sets that are neither open nor closed: the set of rational numbers is a good example. It has no interior points since it contains no open intervals. But it's complement is the collection of irrational numbers, which also contains no intervals, so is not open either.



The whole real line is both open and closed since every point is an interior point (it is in an open interval contained in the real line) and is closed (it contains all its accumulation points..oops, derived points).



The only other subset of the real line that is both open and closed is the empty set, but this is something that needs to be proved. It is essentially the claim that the real line is connected.

Answer 2

Given set A. The set of all derived points is A' (my book called these "boundary points").

The closure of A = cl(A) = A union A'.



A is closed iff A = cl(A). Take the interval [0,1]. A' = {0} union {1} and cl(A) = [0,1] so A is closed.



A is open iff A = cl(A) - A' which means that the set contains no derived points; this is referred to as "the interior of A" and can be denoted int(A). example: the interval (0,1). A' = {0} union {1} but A intersect A' = null so A is open.



Now consider the half-open interval (0,1]. A' = {0} union {1} but A does not equal cl(A) so A is not closed. Also, since int(A) = (0,1) which does not equal A then A is not open. A is neither.



erm....I can't think of an example right now of a set that is both open and closed. I'll have to get back to you on this one.

Answer 3

I'm not sure if I really understand what your question is. Are you asking how we can have sets which are both open and closed?



Here are some examples in the real numbers R:

(0,1) is open.

[0,1] is closed

The empty set and R are the only ones which are both open and closed.

(0, 1] is neither closed nor open



It really depends on how your book defines open and closed, so unfortunately I can't really answer entirely.



In one standard version, here's what you do:

To show a set A is open, show that every point of A is an interior point of A. That is, show that for any point in A, there's an open ball around that point which is entirely contained in A.

To show it's closed, show that if a point is a derived point, that it must be in the set.



That may be no better than your text, though, since most of these are better understood by looking at pictures. If you have any particular question further, I'll be checking back on this question every so often, so just add it as add. info.

Answer 4

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