Sequential Compactness in Metric Spaces: Proving Compactness of Product Metric

[Ted123]

Homework Statement

Suppose $(X,d_X)$ and $(Y,d_Y)$ are sequentially compact metric spaces. Show that $(X\times Y, d_{X\times Y})$ is sequentially compact where $$d_{X\times Y} ((x_1,y_1),(x_2,y_2)) = d_X(x_1,x_2) + d_Y(y_1,y_2)$$ is the product metric.

The Attempt at a Solution

Suppose $(x_n)_{n\in\mathbb{N}}$ is a sequence in $X$ and $(y_n)_{n\in\mathbb{N}}$ is a sequence in $Y$.

Then since $X,Y$ are sequentially compact $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ have convergent subsequences, say $x_{n_k} \to x\in X$ and $y_{n_k} \to y\in Y$ as $k\to\infty$.

It follows that if $(x_n,y_n)_{n\in\mathbb{N}}$ is a sequence in $X\times Y$ with subsequence $(x_{n_k},y_{n_k})_{k\in\mathbb{N}}$ then $(x_{n_k},y_{n_k}) \to (x,y)\in X\times Y$ as $k\to\infty$.

Is this OK so far? Do I now need to show that $(x_{n_k},y_{n_k}) \to (x,y)$ in the metric $d_{X\times Y}$ ?

In which case:

$d_{X\times Y}((x_{n_k} , y_{n_k}),(x,y)) = d_X(x_{n_k} , x) + d_Y(y_{n_k},y) \to 0+0=0$

so $(x_{n_k},y_{n_k}) \to (x,y)$ in the metric $d_{X\times Y}$.

 

[micromass]

Looks ok!