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Is $(\mathbf{N}, d)$ metric space if:

$d(m,n)=\begin{cases} 0& m=n\\ \frac{1}{m}+\frac{1}{n}& m\neq n \end{cases}$. $m, n \in \mathbf{N}$

(Note: it's easily proven that it is.)

If it is, does it genearate discrete topology? What sets do the open balls $B(5,1), B(2,2), B(3,\frac{1}{3})$ represent in that space?

user172906
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2 Answers2

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I presume that you are working in $\mathbb{N}=\left\{ 1,2,3,\dots\right\} $

Singleton $\left\{ n\right\} =\left\{ m\mid d\left(n,m\right)<\frac{1}{n}\right\} $ is open, implying that the topology is discrete.

$B\left(5,1\right)=\left\{ m\mid d\left(5,m\right)<1\right\} =\left\{ 5\right\} \cup\left\{ m\mid\frac{1}{5}+\frac{1}{m}<1\right\} $

So just solve the inequality $\frac{1}{5}+\frac{1}{m}<1$

The others can be done likewise.

drhab
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This space has a nice geometric picture: you can view it as the set of all points $1/n$ for positive integers $n$, with the metric defined as the distance that you must travel between two points going via the origin.

Now the discrete topology means that every singleton set is open, so every singleton must be contained in a ball of some radius. Given this picture, it is straightforward to come up with such a ball for any given $n$: it has radius $1/n$ (or less). So the topology is indeed discrete.

As for the balls you give, the picture makes it clear that $B(5,1)$ contains everything except 1, $B(2,2)$ is the whole space, and $B(3,1/3)$ is the singleton $\{3\}$.

A more interesting space can be obtained by adding the origin itself to the set. It's a good exercise to work out what the open sets look like in this case.

Holographer
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