Showing $A \subseteq \mathbb{R}^p$ is not closed

Question

Give 2 different ways you can go about showing that a set $A \subseteq \mathbb{R}^p$ is not closed in $\mathbb{R}^p$.

I have the following:

We can show that $\mathbb{R}^p \setminus A$ is not open
We can show $A^\prime \not \subseteq A$ , where $A^\prime$ denotes the accumulation point(s) of $A$.

Can anyone please provide me with more ways that once can show it is not closed?

See also http://math.stackexchange.com/questions/501015/a-set-is-closed-if-and-only-if-it-contains-all-its-limit-points. — Dietrich Burde, Apr 09 '15 at 12:58

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

1

There are some other ways.

The closure of $A$ is not $A$ itself. i.e., $\bar A\ne A$

For the case bounded $A$, show that it is not compact. ($A\subset \mathbb R ^{p})$

edited Jun 12 '20 at 10:38

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answered Apr 09 '15 at 13:00

Fermat

1 Answers1