Suppose $(X,d_X)$ and $(Y,d_Y)$ are isometric, then if $X$ is sequentially compact then so is $Y$.
Attempt:
Let $f:Y \rightarrow X$ defines the isometry between two spaces and suppose $(y_n)_{n=0}^\infty$ is a sequence in $Y$. Then $(f(y_n))_{n=0}^\infty$ defines a sequence in $X$. Since $X$ is sequentially compact, there exists a convergent subsequence, say $(f(y_{n_i}))$, that converges to some point $x \in X$. Then by definition of a convergent sequence, $\forall \epsilon>0 ,\exists N\in\mathbb{N},\forall i\in\mathbb{N} (i\geq N \Rightarrow d(f(y_{n_i}), x)<\epsilon)$. Note that as $f$ is an isometry, it is a bijection such that there is some $y \in Y, f(y) = x$ and that we have $\forall \epsilon>0 ,\exists N\in\mathbb{N},\forall i\in\mathbb{N} (i\geq N \Rightarrow d(f(y_{n_i}), f(y))<\epsilon)$. Again as $f$ is an isometry, we will then have
$\forall \epsilon>0 ,\exists N\in\mathbb{N},\forall i\in\mathbb{N} (i\geq N \Rightarrow d(y_{n_i}, y)<\epsilon)$ such that ($y_{n_i}$) is a convergent subsequence. Hence $Y$ is also sequentially compact.
Is my proof correct? Thanks!
Best Answer
Your proof is quite correct.
In fact we can forget about the metric and just assume $f$ is a homeomorphism between spaces $X$ and $Y$.
If $X$ is sequentially compact, let $(y_n)_n$ be a sequence in $Y$ where we have $x_n \in X$ such that $f(x_n)=y_n$ for all $n$.
$(x_n)_n$ has a convergent subsequence $(x_{n_k})_k$ with limit $x$ and by continuity of $f$:
$$f(x)= f(\lim_k x_{n_k}) = \lim_k f(x_{n_k})= \lim y_{n_k}$$ showing that $y_{n_k} \to f(x)\in Y$ as $k \to \infty$. So $Y$, the continuous image of $X$, is also sequentially continuous.
Using the continuous inverse of $f$ we see that $Y$ sequentially compact also implies that $X$ is.
And if you insist on being in a metric setting, sequentially compact is equivalent to compactness there, so also preseved under continuity. Isometries are overkill.