elementary properties of closure Let X be arbitrary subset of R, then $$X\subset \overline X$$
proof by contradiction:
let $x \in X$ and suppose X not a subset of its closure
then for every $y\in X$, $|x-y|> \epsilon $ where $\epsilon>0$
In particular, for $y=x$, $|x-x|>\epsilon \implies 0>\epsilon$ !!contradiction
is the proof logic and writing correct?
 A: The statement $X\subset \overline X$ has a few issues:


*

*If by $\subset$, you mean "proper subset", then the statement is false, since for closed sets $X$, the set $\overline X$ is equal to $X$ so $X$ is not a proper subset. But OK, let's assume $\subset$ means the same as $\subseteq$.

*If you define $\overline X$ as "The smallest closed set that contains $X$", then the statement is trivially true (by definition)

*If you define $\overline X$ as "The intersection of all closed sets containing $X$", which is equivalent to the definition in (2), then proving $x\subset \overline X$ is very easy and does not require $\epsilon$-$\delta$ definitions.


In your case, I am not sure what definition of closure you are using, so please, write down (exactly, not in your own words) what $\overline X$ is.

Edit: You have now pasted your definition of $\overline X$. According to it, 
$$x\in\overline X\iff \forall \epsilon\exists y\in X: |x-y|\leq\epsilon$$
So, in order to prove that $X\subset\overline X$, you must:
  1. Take an arbitrary $x\in X$
  2. Take an arbitrary $\epsilon > 0$
  3. Find a value $y\in X$ such that $|x-y|\leq \epsilon$.
If you want to prove it by contradiction, you must:


*

*Assume that there exists an $x$ for which $x\in X$ and $x\notin \overline X$

*Reach some sort of contradiction which clearly follows from point 1.


You have not performed these two steps. There is no way anyone will understand how you reached your contradiction (I certainly don't). For example, you claim that

if $x\in X$ and $X$ is not a subset of $\overline X$, then for every $y\in X$, the inequality $|x-y|>\epsilon$ is true.

This is simply false in the sense that the statement "$X$ is not a subset of $\overline X$" does not imply that statement.
A: If you're using the definition $\overline{X} = \{x \in \Bbb R\mid \forall \ \epsilon > 0, ]x-\epsilon,x+\epsilon[\cap X \neq \varnothing\}$ (that is, $x$ is a cluster point of $X$), then take $x \in X$ and let any $\epsilon > 0$. Then $]x-\epsilon,x+\epsilon[ \cap X$ contains at least $x$, hence it isn't empty, and $x \in \overline{X}$. The conclusion is that $X \in \overline{X}$. I think that 5xum's answer covers all other possibilities.
