# Compactness of second-countability of $\omega$X$\omega_1$

Please discuss the following properties of the product space consisting of $\omega$X$\omega_1$:

1. Is it compact?
2. Is it 2nd countable?

$\omega$ is the first infinite ordinal and $\omega_1$ is the first uncountable ordinal. The topology is the product topology.

I know that the topological space $[0,\omega_1)$ is not compact, leading me to conclude that the answer for (1) is 'no' using the cartesian product definition of two compact spaces, but here (at least) one of them fails to be compact. Anything pointing me in the right direction will be helpful.

• I do believe the Tychonoff plank is the product of the two closed ordinal spaces $[ 0 , \omega ]$ and $[0 , \omega_1 ]$. (And the deleted Tychonoff plank is obtained by removing $\langle \omega , \omega_1 \rangle$ from it.) – user642796 Dec 8 '14 at 18:14
• I do believe it would be best to split this question into two separate questions. – user642796 Dec 8 '14 at 18:17

HINTS: As Arthur Fischer said in the comments, the deleted Tikhonov plank is not the space that you described, but rather the space obtained by deleting the point $\{\langle\omega,\omega_1\rangle$ from the product space $[0,\omega]\times[0,\omega_1]$.
2. The deleted Tikhonov plank contains a subspace homeomorphic to $[0,\omega_1]$.
Added: It turns out that the space in question is actually $\omega\times\omega_1$, so you can simply use the fact that the space contains a clopen subspace homeomorphic to $\omega_1$ to answer both questions.
• Hi thanks for the hints, but I got mixed up with the Tychonoff plank shown in the lecture notes with the example I'm supposed to show properties for. So in this one, I am supposed to answer whether $\omega$X$\omega_1$ is compact and second second countable. I know that $\omega_1$ is not compact so using the fact that the product of two compact spaces must be again compact I can provide the counterexample. Same can be said for $\omega_1$ not being second countable. Please let me know if I am in the right direction given this correction. – LordVader007 Dec 10 '14 at 6:28
• @LordVader007: It’s true that $\omega\times\omega_1$ is not compact, but you can’t use the product theorem to show it. Using the fact that $\omega$ is a discrete space, you can easily construct an open cover with no finite subcover. Alternatively, you can find a closed subset of $\omega\times\omega_1$ that is homeomorphic to the non-compact space $\omega_1$, and conclude that $\omega\times\omega_1$ is not compact (why?). Or you could find a closed subspace homeomorphic to $\omega$ and draw the same conclusion. You can also find an open subspace homeomorphic to $\omega_1$ and use it to ... – Brian M. Scott Dec 10 '14 at 6:47
• ... conclude that $\omega\times\omega_1$ is not second countable, because ... ? – Brian M. Scott Dec 10 '14 at 6:48