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Let $d_1$, $d_2:X \times X \to [0, \infty)$ be any of two metrics on a set $X$ and $\{x_n\}_{n \in \mathbb{N}}$ be a sequence in $X$ that converges to $x_* \in X$ in $(X, d_1)$ and $y_* \in X$ in $(X, d_2)$. Then can we conclude that $x_* = y_*$?

If $d_1$ and $d_2$ are equivalent metrics, then $x_* = y_*$ is true (here is a discussion and proof when $d_1$ and $d_2$ are strongly equivalent, and here when $d_1$ and $d_2$ are equivalent). But I cannot figure out it still holds for any arbitrary metrics $d_1$ and $d_2$.

  1. Can you help me prove or disprove that $x_* = y_*$? (Maybe there is a counter example of $x_* = y_*$, but I failed to find it).

  2. If $x_* = y_*$ is not true in general, is there some weaker condition or notion than the equivalence of $d_1$ and $d_2$ that guarantees $x_* = y_*$?

Many thanks in advance for your help.

Community
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Jae Young Lee
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1 Answers1

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Consider $X = \mathbb{R}$, $d_1(x,y) = \lvert x-y\rvert$ and $d_2(x,y) = \lvert f(x) - f(y)\rvert$ where $f\colon \mathbb{R} \to \mathbb{R}$ is given by

$$f(x) = \begin{cases} 0 &, x = 1 \\ 1 &, x = 0 \\ x &, x \notin \{0,1\}.\end{cases}$$

Let $x_n = 2^{-n}$. Then $x_n \to 0$ with respect to $d_1$, but $x_n \to 1$ with respect to $d_2$. It should be clear how to construct similar examples on arbitrary infinite $X$.

There is indeed a weaker condition than equivalence implying that the limit of a sequence with respect to both metrics is the same, provided the sequence converges with respect to both metrics. This is already the case if the two metrics are comparable, that is, the topologies defined by the two metrics are comparable. For if a sequence converges in a topology, it converges to the same limit in every coarser topology.

Daniel Fischer
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