Does being Cauchy in $M$ imply being Cauchy in a subspace?

Question

I've just seen an example about the sequence $x_n = \frac{1}{n}$ being Cauchy in $(0,1]$ because "Cauchy in $\mathbb{R}$ $\implies$ Cauchy in $(0,1]$".

Is this true for a general metric space $M$ and some subset $X \subset M$?

Intuitively this seems like it should "obviously" be true.

Answer 1

Let $(M, d_M)$ be a metric space, let $X \subset M$, let $(x_n)_{x\in\mathbb{N}}$ be a Cauchy sequence in $M$ with respect to $d_M$, let $x_n\in X$ for all $n\in\mathbb{N}$, and let $d_X$ be the subspace metric of $d_M$ with respect to $X$.

Then since $d_X(x_i, x_j) = d_M(x_i, x_j)\ \ \forall i, j$ it follows that $(x_n)_{x\in\mathbb{N}}$ is a Cauchy sequence in $X$ with respect to $d_X$.

It also follows from this and the fact that any subsequence of a Cauchy sequence is Cauchy that for any cauchy sequence, the subsequence induced by a metric subspace is also Cauchy, if it exists.