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Exercise If $X,Y $are two topological spaces and $X$ is not compact then, $Y$ is any space then $X \times Y$ is not compact.

My attempt: So we know that topological space is compact if every open cover has a finite subcover, so take an open cover of $X \times Y$. Say $U_\alpha$. We know that each open set in $X\times Y$ is of the form $U_i = V \times V'$, where V is open in X and V' is open in Y. Hence $U_\alpha = V_\alpha \times V'_\alpha$ ,where $V_\alpha $ and $V'_\alpha$ are open covers of X and Y respectively. But since X and Y are not compact we know that there does not exist finite subcover of neither $V_\alpha $ nor $V'_\alpha$. Hence we can not find finite subcover of $V_\alpha \times V'_\alpha$ and hence for $U_\alpha$.

Is it enough or it is not sufficient?

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    $\begingroup$ This is a product, not cross product. Cross product is a pretty specific operation on vectors, not general topological spaces. I fixed your title for you. Modulo a couple of minor issues (you don't know anything about $Y$), this is effectively correct. I would fix it to say "Since $X$ is not compact, we know that there is no finite subcover, so we cannot have a finite subcover for $X\times Y$ and so it is no compact." $\endgroup$ – Cameron Williams Jun 2 '18 at 16:12
  • $\begingroup$ You should note that only $X$ is given as not compact. $\endgroup$ – Mark Bennet Jun 2 '18 at 16:15
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    $\begingroup$ To show that $X \times Y$ is not compact, you must exhibit an open cover that does not have a finite subcover. Moreover, it is not the case that every open cover of a noncompact set does not have a finite subcover; for example the cover $\{X\}$ of $X$ is a finite subcover of itself, regardless of whether $X$ is compact or not. Lastly, not every open set of $X \times Y$ is of the form $V \times V'$ $\endgroup$ – Chaitanya Tappu Jun 2 '18 at 16:20
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If $X \times Y$ were compact, then $X = \pi_X[X \times Y]$ is also compact, as the continuous image of a compact space under the continuous projection map $\pi_X: X \times Y \to X$ defined by $\pi_X(x,y) = x$.

Using covers: let $U_\alpha, \alpha \in A$ be an open cover of $X$ without a finite subcover.

Then $\{U_\alpha \times Y: \alpha \in A\}$ is an open cover of $X \times Y$ without a finite subcover (if $Y\neq \emptyset$). So $X \times Y$ is not compact.

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  • $\begingroup$ Empty Y needs to be excepted. $\endgroup$ – William Elliot Jun 2 '18 at 22:56
  • $\begingroup$ +1 for projection map method. It may be instructive for OP to work through the open cover method since its instructive, but when proving a property of some object I try to always work externally $\endgroup$ – Prince M Jun 4 '18 at 1:54
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"We know that each open in $X\times Y$ is of the form..." This isn't true. The open unit disc $\{x: |x|<1\}$ is open in $\Bbb R\times\Bbb R$, for instance.

Hint: If you want to prove compactness, you need to take an arbitrary cover. If you want to disprove compactness, you can choose a cover to your liking. Use the non-compactness of $X$ to construct a cover of $X\times Y$ such that...

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  • $\begingroup$ So if we take an open cover of X that does not have finite subcover and and open cover of Y. It is true that it is a cover of $X \times Y$ right? Ant the since there is no finite subcover for open cover of X consequently there is no finite subcover of a produc.. Is that true? $\endgroup$ – user557550 Jun 2 '18 at 16:36
  • $\begingroup$ @MariyaKav Or simpler, let $\{U_\alpha\}$ be such a cover of $X$. Then $\{U_\alpha\times Y\}$ is such a cover of $Y$. In other words, do what you said, but chose, explicitly, the simplest open cover of $Y$. $\endgroup$ – Arthur Jun 2 '18 at 16:48
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It is not true that every open set in $X\times Y$ is of the form $V\times V'$; for the easiest example, take the unit disk in $\mathbb R^2$.

Also, you won't get anywhere by starting with some open cover. You want to show that something is not compact, you need to exhibit a single open cover that does not admit a finite subcover. So, using that $X$ is not compact, you know that there exists an open cover with no finite subcover; now use this information to construct an open cover of $X\times Y$ that does not admit an open subcover.

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  • $\begingroup$ So if we take an open cover of X that does not have finite subcover and and open cover of Y. It is true that it is a cover of $X \times Y$ right? Ant the since there is no finite subcover for open cover of X consequently there is no finite subcover of a produc.. Is that true? $\endgroup$ – user557550 Jun 2 '18 at 16:33
  • $\begingroup$ Yes. $ \ \ \ \ $ $\endgroup$ – Martin Argerami Jun 2 '18 at 16:43

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