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Sequential Compactness

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    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.

    3. 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].
    Last edited: Nov 22, 2011
  2. jcsd
  3. Nov 22, 2011 #2
    Looks ok!!
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