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!