# Discrete and metric topologies equivalence

Given a set $X$, define a function $d:X\times X\rightarrow \mathbb{R}$ by $d(x,y) = 1$ if $x\neq y$ and $d(x,y)=0$ if $x=y$. Show that the metric topology on $X$ is equal to the discrete topology.

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Can you add the (homework) tag please, and maybe say something about what you've tried in solving this? –  Guess who it is. Apr 30 '11 at 1:13
What a discrete hint! –  ncmathsadist Apr 30 '11 at 2:51
The title does not reflect the question. –  lhf Apr 30 '11 at 3:00
$d$ should take values in $\mathbb R$, not $\mathbb R^2$. –  lhf Apr 30 '11 at 3:01

Hint: What does the ball of radius $1/2$ around $x$ look like?

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To show that two topologies are equivalent, you want to show that they have the same open sets. Remember that in a metric space, the open sets are defined to be those for which, given any point inside the set, you can draw an open ball around that point and remain within the set. Theo's hint above should help you to see what the open sets in the metric topology look like. Every set in the discrete topology is open (by definition), so you want to show that every set in the metric topology is open, too. –  Anon445 Apr 30 '11 at 12:44
I could use some more understanding concerning this question... I would like to keep it going, even if I am not the OP. As for the hint, if x = y then $x^2 + y^2 = 2x^2$ so if we originally had $r^2 = (1/2)^2 = (x^2 + y^2)$ then we would have, by x = y substitution, $1/2 = x\sqrt{2}$. What does this show though? –  Relative0 Sep 29 '13 at 19:12