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Let $(M,\tau)$ be a metric space and $(Y,\tau_{Y})$ a subspace of a the metric space with an equivalent metric.

Prove: for all $\{x_{n}\}\subseteq Y$ and $x\in Y$, $x_{n}\rightarrow x$ according to $\tau\iff x_{n}\rightarrow x$ according to $\tau_{Y}$

The given proof is: if $x_{n},x\in Y$ so $\tau(x_{n},x)=\tau_{Y}(x_{n},x)$ and by the definition of equivalent of metrics $\tau(x_{n},x)\rightarrow 0\iff\tau_{Y}(x_{n},x)\rightarrow 0$

What does it follow from the "definition of equivalent of metrics"?

because in the case that all of the elements are form the subspace of the metric space both $\tau$ and $\tau_{Y}$ "behave" the same?

what if we only knew that $\tau(x_{n},x)\rightarrow 0$ when $x_{n},x\in M$ and $x_{n},x\notin Y$, can we still say that $\tau_{Y}(x_{n},x)\rightarrow 0$?

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gbox
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    For you third question the answer is no. $\tau_{Y}$ is a metric defined on $Y$, thus $\tau_{Y}(x_n,x)$ is undefined for $x_n$,$x$ not in $Y$ – Uskebasi May 07 '17 at 13:44
  • @QWERTZ thanks that is what I thought – gbox May 07 '17 at 14:00
  • Now, I'm not sure what definitions you have but if it is given that $\tau(x_{n},x)=\tau_{Y}(x_{n},x)$ you don't need equivalence of metrics (or to be more precise, the two metrics are trivially equivalent and it is not necessary to state it explicitely, since you don't need it for the proof). I think the statement can be proved even if the metrics are equivalent, i.e. without this equality $\tau(x_{n},x)=\tau_{Y}(x_{n},x)$ (which is more restictive than equivalence) – Uskebasi May 07 '17 at 14:10

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