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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.

Sahiba Arora
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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.

hmakholm left over Monica
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    That's a great example and exactly the kind of thing I wanted. Thanks. – Sahiba Arora Oct 01 '15 at 13:19
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    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...) –  Oct 01 '15 at 14:13
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    @Pakk It is just the usual metric on $\mathbb{R}$, with 0 and $\pi$ swapped. – Yakk Oct 01 '15 at 15:44
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    @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. – hmakholm left over Monica Oct 01 '15 at 18:22