Every subset of discrete space is open and closed.

Question

Had a look at the solutions and their answer is different. Was wondering if my attempt works and would love some feedback? Thank you in advance!

Question:

Let $X$ be a non-empty set equipped with the discrete metric. Prove that every subset $Y$ of $X$ is both open and closed.

My attempt:

Let $r = 2$

Since we are dealing with the discrete metric, $∀a,b∈X, d(a,b)≤1$

$⇒ B(y,2) = \{z∈X|d(y,z)≤1<r = 2\}$ and therefore $∃r>0$ such that $B(y,r)⊆X$

Therefore any subset $Y$ is open $⇒$ any subset $X\setminus Y$ is closed. Therefore any subset $Y⊆X$ is both open and closed.

For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — Henrik supports the community, Dec 13 '23 at 14:55
Solution verification tag is not for verifying your solutions, but particular steps of confusion in your solution. That's why I'm voting to close this question as a duplicate. — Jakobian, Dec 13 '23 at 15:01
Does this answer your question? Show that in a discrete metric space, every subset is both open and closed. — Jakobian, Dec 13 '23 at 15:02
It gives me an answer but I'm still unsure of whether my proof is correct, since it goes in a very different route from most solutions I can find, where they set r<1 whereas I am attempting it by setting r>1. This is the step in particular that I am confused about. — Thatbf1nub, Dec 13 '23 at 15:19
Well, ${z\in X : d(y, z) \leq 1 < r = 2}$ while not incorrect, you rarely write things like that and I want to point out there might be some confusion on your part on how to write sets. Your solution is wrong but if you take $B(y, 1/2) = {y}$ then you'll be able to show that $Y$ is open. — Jakobian, Dec 13 '23 at 15:48
$B(y, 2) = X$ so its rarely contained in $Y\subseteq X$ unless $Y = X$ — Jakobian, Dec 13 '23 at 15:49
@Jakobian just revisited this question and realised I had copied the definition of an open set wrong hence the confusion. Your answer helped me realise this so thanks a lot for your help! — Thatbf1nub, Dec 13 '23 at 19:59
No problem. You can click on "close" and choose the question this is a duplicate of if you refused the suggested close vote for a duplicate. — Jakobian, Dec 13 '23 at 20:17

Every subset of discrete space is open and closed.

0 Answers0