Simple question about discrete metric and openness.

Question

You may think this is silly question, but I'm really confused.

In discrete metric, every singleton is an open set.

And, the proof goes like this

$\forall x \in X$, by choosing $\epsilon < 1$, $N_\epsilon(x) \subset ${$x$}

However, if discrete metric is defined in $X$,

How can we use the radius less than 1 when we only have 0 and 1?

Because the distance is a function defined to R+, not only the image — Martín Vacas Vignolo, Mar 24 '16 at 15:13
In fact, every subset of $X$ is an open set, since by the same argument you can find an open ball within the subset for every point in the subset. This is the "discrete topology". — Tom Collinge, Mar 24 '16 at 15:23

score 2 · Accepted Answer · answered Mar 24 '16 at 15:16

2

A metric on $X$ is a mapping $d:X \times X \to [0,\infty[$, so you can take $d = 1/2$ even if no pairs with distance $1/2$ exist.

answered Mar 24 '16 at 15:16

Henricus V.

18,694

score 0 · Answer 2 · answered Mar 24 '16 at 15:35

0

You don't need to take $\varepsilon<1$, you may take $\varepsilon=1$ also (can you see the reason?). But you need to know that the range of the metric (here $d$) is $\mathbb{R}^+$.

answered Mar 24 '16 at 15:35

Simple question about discrete metric and openness.

2 Answers2