1. Mar 18, 2012

cragar

1. The problem statement, all variables and given/known data
If I just had the set containing $\pi$ on the real line.
So this is an isolated point. Is this set closed?
3. The attempt at a solution
I think this set is closed because it contains its limit points, because it only has one point.

2. Mar 18, 2012

NewtonianAlch

Hmm, that's an interesting question. I guess it would be closed because that's the only point in the set, and there's no option for an e-neighbourhood around it.

3. Mar 18, 2012

Dick

I think you are thinking about it correctly but that you are even asking this question makes me wonder. Can you explain your doubts further? What's the exact definition of a limit point?

4. Mar 18, 2012

cragar

thats what I thought too

5. Mar 18, 2012

zooxanthellae

If we wanted to put things on more definite footing, we could assume that it isn't closed. Then there is some point x1 such that for all E > 0 there is some x in our set such that d(x,x1) < E. However, since x = π in all cases, then the only such point is π, a contradiction.