# Basic topology - Limit points and closure

1. Sep 26, 2012

### ran13

This isn't really hw, just me being confused over some examples.

I have 'learned' the basic definitions of neighborhood, limit point, closed, and closure but have some trouble accepting the following examples.

1. For Q in R, Q is not closed. The set of all limit points of Q is R, so its closure is R.

Between every two real numbers is a rational, I know as fact. But can't it be that between any two rational numbers is another rational?

- Since we can adjust the neighborhood of any rational to be of any radius, won't we cover all rational numbers in this way? So is it incorrect to say that the limit points of Q can be Q itself?

- Or is it simply a matter of being as complete as possible? Not complete in the real analysis sense, but in the sense that possibilities are included. It seems more complete to account for any point that can be a limit point of Q. In this case, I know it's correct to say more generally that R is the set of all possible limit points of Q, since Q is in R.

2. E = { 1/n | n = 1, 2, 3, ... }

The limit point of E in this case is 0, so the closure of E is E U {0}.

This is the set of rationals in (0,1] of the form 1/n. So 1 cannot be a limit point since we can chose the neighborhood of 1 to be, say 0.1, and it will not include any other points of E.

- Is this the same reasoning behind saying that no other 1/n in E can be a limit point? And the same reasoning behind saying that [0,1] in R is not the set of all limit points of E?

- But why then is zero a limit point? Can't we adjust its neighborhood so that it does not cover a number of the form 1/n? Are there no non-reducible rationals that are very close to zero that don't have the form 1/n, some tiny number over a large number?

It seems like the best form of such a number would be of form 1/n, but... Do all rationals near zero have the form 1/n? This is what the example seems to be saying since we can pick any neighborhood, yet 0 is still a limit point.

Any clarity would be appreciated... Pardon my slowness but I want to know what mistakes I'm making here. Thanks :)

2. Sep 26, 2012

### jbunniii

Correct.

Yes, if p and q are rational, then so is (p+q)/2, which is midway between p and q.

If you are asking whether every point of Q is a limit point of Q, the answer is yes.

Correct, every point of Q is a limit point of Q, but those are not the only limit points of Q. Every real number is a limit point of Q.

Correct.

Choose any neighborhood of zero with positive radius R > 0. No matter how small R is, you can find an N such that 1/N < R. Thus 0 is a limit point of the set.

No, but they don't have to have that form. The definition doesn't require that EVERY point of every neighborhood of zero is a point of the form 1/N. It just requires that every neighborhood of zero must contain at least one point of that form. An easy argument shows that this in fact means that each neighborhood of zero must contain infinitely many such points, but that is not the same as saying that all points in the neighborhood must be of that type.

3. Oct 7, 2012

### ran13

Thank you for the help. I did get around to doing the proof that every neighborhood of zero contains infinitely many points in form 1/n, though it did take a while... Thanks again.

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