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Sequential Compactness

  • Thread starter Ted123
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Homework Statement



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

The Attempt at a Solution



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

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

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

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

In which case:

[itex]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[/itex]

so [itex](x_{n_k},y_{n_k}) \to (x,y)[/itex] in the metric [itex]d_{X\times Y}[/itex].
 
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Answers and Replies

  • #2
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Looks ok!!
 

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