# Does there exist a metric on $\mathbb{R}$ such that every sequence converges?

A metric $$d:\mathbb{R} \times \mathbb{R} \to [0,\infty)$$ is a function satisfying the following three properties:

1. $$d(x,y)\geq 0$$ and $$d(x,y)=0 \iff x=y$$ (Positivity)

2. $$d(x,y)=d(y,x)$$ (Symmetry)

3. $$d(x,y)\leq d(x,z)+d(z,y)$$ (Triangle Inequality)

A sequence $$(x_n)$$ is said to converge to a limit $$x\in \mathbb{R}$$ if $$\forall \epsilon >0, \exists N\in \mathbb{N}, \forall n\geq N, \implies d(x_n,x)<\epsilon$$.

If no such $$x\in\mathbb{R}$$ exists, then the sequence is said to diverge.

Some examples of metrics include: the Euclidean metric, discrete metric, and $$2$$-adic metric.

Question: Does there exist a metric on the real numbers such that every sequence converges to some real number?

• What about $x_n=(-1)^n$? Mar 1, 2023 at 8:32

If $$\{x,y,x,y,\cdots\}$$ converges then it must be Cauchy which implies that $$d(x,y)=0$$ for all $$x,y$$. So there is no such metric not only on $$\mathbb R$$ but also on any set with more than one point.

First of all, if a sequence is eventually constant then it is convergent to this eventually constant value. This holds even in topological spaces.

Secondly, if a sequence is convergent to some limit then its every subsequence is also convergent to the same limit. Again, this hold for all topological spaces.

Now assume that $$X$$ is a Hausdorff space (which covers all metric spaces). This assumption gives us an important property: a convergent sequenece has a single, unique limit.

Let $$x,y\in X$$ and define

$$v_n=\begin{cases} x&\text{if }n\text{ is even} \\ y&\text{otherwise} \end{cases}$$

So now we have an assumption that $$v_n$$ converges, to say $$v$$. But the constant $$x$$ is a subsequence of $$v_n$$ converging to $$x$$. By the uniquness of limit $$v=x$$. Analogously $$v=y$$, and therefore $$x=y$$. And therefore we've shown that such situation can only happen when $$X=\{x\}$$ has a single point to begin with.

Note that if you are not familiar with topological spaces, then that's not an issue: the above works the same for metric spaces.

Finally, note that such thing can happen when Hausdorff is not assumed. For example any set $$X$$ with anti-discrete topology $$\{\emptyset, X\}$$ will have every sequence convergent. Or more generally a space with a point $$x_0\in X$$ such that $$X$$ is the only open neighbourhood of $$x_0$$. It is an interesting question how to classify such spaces. I don't know the answer.

• One probably has to exclude the (trivial) case that $X$ has less than 2 elements. Or is that not possible in a Hausdorff space? Mar 1, 2023 at 9:04