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I read that the product of two countably compact spaces isn't countably compact.

My question is that since sequentially compact implies countably compact, does this statement imply that the product of a sequentially compact space and a countably compact space is not countably compact?

I'm not sure how to go about proving whether or not it is. I know that sequentially compact means every sequence has a convergent subsequence, and countably compact means every sequence has a cluster point.

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Yes.

Let $X$ be countably compact and $Y$ be sequentially compact, then $X\times Y$ is countably compact.

Proof: Cosider a countably infinite set $A=\{z_1,z_2,...\}\subset X\times Y$, where $z_i=(x_i,y_i)$ for $i=1,2,...$ and $z_i\not=z_j$ whenever $i\not=j$, and let $y_{k_1},y_{k_2},...$ be a subsequence of $y_1,y_2,...$ that converges to a point $y\in Y$. If the set $\{x_{k_1},x_{k_2},...\}$ is finite, then there exists a point $x \in X$ and a subsequence $k_{m_1},k_{m_2},...$ of the sequence $k_1,k_2,...$ such that $x_{k_{m_i}}=x$ for $i=1,2,...$; if the set $\{x_{k_1},x_{k_2},...\}$ is infinite, then it has an accumulation point of $x\in X$. One readily sees that in both cases $(x,y)\in X\times Y$ is an accumulation point of the set $A$.

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  • $\begingroup$ Thanks @Paul, but I don't follow why the set $\{x_{k_1},x_{k_2},\ldots\}$ could be finite. My understanding of a sequence is that it is infinite, so if we're picking a subsequence $\{y_{k_i}\}$ and we get another subsequence $\{x_{k_i}\}$ for free, then the set $\{x_{k_1},x_{k_2},\ldots\}$ should be infinite since it corresponds to a sequence. $\endgroup$ – Polly Nomial May 4 '13 at 14:36
  • $\begingroup$ @PollyNomial: If the elements of the set $\{x_{k_1}, x_{k_2}, ...\}$ are different for each other, then it is infinite. However, some elements of this set may be the same. $\endgroup$ – Paul May 5 '13 at 11:30

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