Determining what sequences converge in a given metric space?

Question

I was given a space such that $\forall x,y \in \mathbb{R}$, we set $\delta(x,y) = 0$ if $x = y$, and $\delta(x,y) = 1$ if $x \neq y$. I'll refer to this metric space as $X$.

My task is then determining which sequences converge in this metric space. I've proved that it is a metric space, but now I'm wondering how I should think about considering which sequences converge in the space?

My initial thought is that I would need sequences which satisfy $\delta(x,x_n) < \epsilon$ whenever $n \geq N \in X$. I know I am basically setting a metric of all numbers such that the distance between them is either 1 or 0. All sequences which converge, given our definition, would have $\delta(x,x_n) = 0$ I would think, considering that they have to be less than $\epsilon$. But then also all sequences which converge a distance of 1 from their considered $x$? I'm not sure.

Anyways, any help would be greatly appreciated!

Answer 1

This metric is known as the discrete metric. Firstly, what does it mean if a sequence does not converge in a metric space? For all $x\in X$, there exists an $\varepsilon>0$ such that for all $N\in\mathbb N$, there exists an $n\ge N$ such that $d(x_n,x)\ge \varepsilon$.

It should be clear that the only sequences that converge with respect to the discrete metric are the sequences that are eventually constant, since distinct points are all distance $1$ apart. That is $x_n = x$ for all $n\ge N$ for some sufficiently large $N$ and a fixed $x\in X$.

Let's prove that. If $\{x_n\}$ is a sequence that is not eventually constant, then let $x\in X$ be arbitrary and set $\varepsilon=1/2$. Let $N\in\mathbb N$ be arbitrary. In fact, $d(x_n,x)=1>1/2=\varepsilon$ for infinitely many $n\ge N$, since $x_n$ is not eventually constant. Since $N$ was arbitrary, we're done by our opening remarks about the definition of not converging.

Answer 2

As you correctly realized $\delta(x,x_n)<\varepsilon$ for any $\varepsilon>0$ and all $n$ greater than some $N$. Thus, $\delta(x,x_n)=0$ for almost all $n$ (choose $\varepsilon=\frac 12$) which implies that any convergent sequence must become stationary at some point.