- #1

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter dustbin
- Start date

- #1

- #2

clamtrox

- 938

- 9

There is an easier way to prove this too though. Do you know that the closure of S is the intersection of all closed subsets containing S?

- #3

dustbin

- 240

- 5

- #4

Dick

Science Advisor

Homework Helper

- 26,263

- 621

Your original proof looks fine to me. Except when you pick an epsilon, it should be "there exists epsilon" not "for all epsilon".

Share:

- Last Post

- Replies
- 18

- Views
- 653

- Last Post

- Replies
- 4

- Views
- 548

- Replies
- 3

- Views
- 362

- Replies
- 5

- Views
- 380

- Replies
- 2

- Views
- 287

- Replies
- 7

- Views
- 478

- Replies
- 6

- Views
- 879

- Replies
- 43

- Views
- 2K

- Replies
- 7

- Views
- 1K

- Replies
- 16

- Views
- 339