Interior, boundary, and closure of sets.

Question

In the usual metric on $ \Bbb{R} $ , find the interior, boundary, and closure of the following sets:

$A = (1,2]$

$B = \Bbb{N}$

$C = \Bbb{Q}$

For $A, I$ got the interior to be $(1,2)$, and the closure to be $[1,2]$. I am unsure how to express the boundary.

For $B, I$ got the interior and closure to be $\Bbb{N}$, and I am not sure how to express the boundary.

Not sure of any of these for $C$.

so for A: int = (1,2), boundary = {1, 2}, closure = [1,2] ? – kt046172 Oct 13 '19 at 20:20 — kt046172, Oct 13 '19 at 20:20
for B: int = {}, boundary and closure= $\Bbb{N}$ ? – kt046172 Oct 13 '19 at 20:21 — kt046172, Oct 13 '19 at 20:21
and for C: int and closure = $\Bbb{R}$, boundary = {} ? – kt046172 Oct 13 '19 at 20:22 — kt046172, Oct 13 '19 at 20:22

Fareed Abi Farraj · Answer 1 · 2019-10-14T03:53:37.530

Yes you're right about $A$.

For the set $B=\Bbb N$ we have

$\overline {\Bbb N}=\Bbb N$ since if it is not true then there exist a point $x \in \overline {\Bbb N} \setminus \Bbb N$ but if so, then any ball of center $x$ must intersect $\Bbb N$, i.e. $\forall r>0 B(x,r) \cap \Bbb N \ne \emptyset$ but we have $\exists n\in \Bbb N$ such that $n<x<n+1$, now take for example $r=\frac{max\{x-n,n+1-x\}}{4}$ now $B(x,r) \cap \Bbb N=\emptyset$ since there is no integer in the ball $B(x,r)$ and this is our contradiction.

And also similarly $\overline {\Bbb Q}=\Bbb R$

The interiors of $B$ and $C$ are empty, can you prove this?

Interior, boundary, and closure of sets.

1 Answers1