# what is the interior and the closure of $\{4\}$ and of $\{2,4,6,8…\}$?

let $$n \in \mathbb{N}$$ . Set $$A_n = \{ n , n+1, n+2,....\}$$ and $$T = \{ \emptyset , A_n\}_{ n\in \mathbb{N} }$$

$$1.$$Is $$T$$is separated ?

$$2.$$ what are dense set in $$T$$ ?

$$3.$$ what is the interior and the closure of $$\{4\}$$ and of $$\{2,4,6,8.....\}$$?

My attempt : for $$1,$$ I know that $$T$$ is not separated because take $$A_1 = \{ 1, 2,3,...\}$$ and $$A_2 = \{ 2, 3,4,,,,,,,\}$$ here both $$A_1 \cap A_2 \ne \emptyset$$

For $$2)$$ i thinks $$\mathbb{N}$$

For $$3)$$ im confused

Any hints/solution will be appreciated

thanks u

• Please use \emptyset to denote the empty set, not $\phi$! – B.Swan May 1 at 10:47
• @B.Swan okkss.. – jasmine May 1 at 10:48

## 2 Answers

$$1)$$ Actually, $$A_n\cap A_m\neq \emptyset$$ for every $$n,m\in\Bbb{N}$$. That should help you prove that no pair of points can be separated.

$$2)$$ A subspace $$S$$ is dense when its closure $$\overline{S}$$ is the whole space. A point $$x\in\overline{S}$$ is a point for which every neighbourhood contains a point of $$S$$. It is not hard to prove that $$S$$ is dense if and only if $$S$$ is infinite. If $$S$$ is finite, then it has a maximum element $$s_0$$, so any point $$s>s_0$$ has a neighbourhood which does not intersect $$S$$. Similarly, if for every $$n\in \Bbb{N}$$ there is a neighbourhood containing a point in $$S$$, then $$S$$ must be infinite.

$$3)$$ Their interior is empty because no open set is contained in $$\{4\}$$ nor in $$\{2,4,6,\dots\}$$.

• thanks u javi,, – jasmine May 1 at 11:09

Your answers:

1) Incorrect. The fact that two particular open sets intersect does not imply at all that the topology is not separated. You should pick two natural numbers $$m,n$$ (call $$m$$ the lesser one, for example) and try to show that this points can not be separated.

2) Incorrect. Of course $$\Bbb N$$ is dense. But how are the closed sets? What would be the closure of the set of, say, even numbers?

For the third one, perhaps you have a clearer idea of how the topology is. Good try!

• .okss @ajotatxe for 2) i thinks ur assumption is wrong why are r u taking even number ? for third i think interior of$\{4\}= \emptyset$ and closure $\{4\} = \mathbb{N}$ – jasmine May 1 at 11:01
• @jasmine The closure of $\{4\}$ is $\{1, 2, 3, 4\}$. Try showing that, if an open set contains $1$, $2$, or $3$, then it must contain $4$. Also, if $n > 4$, then try finding an open set that contains $n$ but not $4$. (Or just write out all the closed sets, and find the smallest one that contains $4$.) – Theo Bendit May 1 at 11:06
• @TheoBendit thanks u – jasmine May 1 at 11:09