# Sequence converging to different limits

In a metric space, is it possible to find a sequence which converges to two different limits with respect to two different metrics?

Obviously the metrics can't be equivalent.

• What if we take one ae usual metric and the other one on the lines d(x,y) =1 for $x \ne y$ in the second one, the sequence will eventually be constant Oct 1, 2015 at 12:49
• @Shailesh, but they won't converge to two different limits, then, right? Discrete metric won't ever work. Oct 1, 2015 at 13:16
• Yes, I realised that after posting the comment. Thanks Oct 1, 2015 at 13:34

With two different metrics? Yes, obviously. (But with two different metrics it is not the same metric space, by definition -- the concept of a metric space includes which metric we're using).

For example take $\mathbb R$ with respectively the standard metric, and the metric $$d_2(a,b)=|f(a)-f(b)| \quad\text{where }f(x)=\begin{cases} \pi & \text{if }x=0 \\ 0 & \text{if }x=\pi \\ x & \text{otherwise} \end{cases}$$

Then $a_n=\frac1n$ converges to $0$ in the usual metric but to $\pi$ in the $d_2$ metric.

• That's a great example and exactly the kind of thing I wanted. Thanks. Oct 1, 2015 at 13:19
• Is there a name for the type of metric that $d_2$ is? (At first I thought this could not be a metric, but it surprised me to see that it really is...)
– user193810
Oct 1, 2015 at 14:13
• @Pakk It is just the usual metric on $\mathbb{R}$, with 0 and $\pi$ swapped.
– Yakk
Oct 1, 2015 at 15:44
• @Pakk: Whenever $(X,d_X)$ is a metric space and $f:Y\to X$ is an injection, you can make $Y$ into a metric space by defining $d_Y(a,b)=d_X(f(a),f(b))$. In the particular case where $Y\subseteq X$ and $f$ is the inclusion map, this gives the subspace metric, but I don't recall seeing any nice name for the general case. One might say something like "the metric on $Y$ induced by $f$", though. Oct 1, 2015 at 18:22