Equality with metric in metric spaces

Question

Claim

Suppose that $(X,d)$ is a metric space. Then $\forall x,y \in X$, $$x=y \iff d(x,y) < \epsilon$$ for all $\epsilon >0$.

Proof:

Proving $\Rightarrow$

Suppose $x, y \in X$ are arbitrary. Then, if $x=y$, then by definition of the metric $d$, $$d(x,y)=0 < \epsilon$$

Proving $\Leftarrow$

I will prove this by contradiction. Suppose that $x \ne y$ and $\forall \epsilon > 0$, $$d(x,y) < \epsilon$$

By definition of $d$, $$ d(x,y) > 0$$

Since $d(x,y) > 0$, take $\epsilon = d(x,y)$. Now, $$0 < d(x,y) < \epsilon = d(x,y)$$

But, we have $d(x,y) < d(x,y)$. Hence, a contradiction.

Question

Is this proof valid and correct? Also, this question may seem out of topic, but is it true even if "$=$" is replaced by "$\sim$", an equivalence relation? I think it is true when it is replaced with "$\sim$" because $d$ is reflexive and symmetric when defining the relation $x \sim y :\iff d(x,y)<\epsilon$, for all $\epsilon > 0$. However, I don't know about the transitivity part. Is it

Suppose that $x, y$ and $z \in X$ such that $d(x,y)< \frac{\epsilon}{2}$ and $d(y,z)< \frac{\epsilon}{2}$. Then by the Triangle Inequality definition of the metric, $$d(x,z) \leq d(x,y) + d(y,z) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$

?

Answer 1

You first part of the $\impliedby$ proof has a mistake. You write

"suppose $x < y$"

but you should have written

"suppose $x \ne y$".

You've also been a little slopping elsewhere, saying "by definition of $d$"; the right thing to say, in the first use (in the $\implies$ part), would be

"then by the first property of a metric, $d(x, y) = 0$"

I'm assuming here that you've defined a metric to be a function to the nonnegative reals with three properties, something like this:

1. for all $x$, $d(x, x) = 0$
2. for all $x, y$, $d(x, y) > 0$ if $x \ne y$.
3. [triangle inequality]

In the second use, I'd replace "By definition of $d$" with "By the second property of a metric".

Finally, I'd consider thinking about how to prove the $\impliedby$ case without using contradiction, just because it's good practice.