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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?

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    $\begingroup$ What about $x_n=(-1)^n$? $\endgroup$
    – Didier
    Mar 1, 2023 at 8:32

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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.

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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.

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

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