Is $[0,1) \times [0,1)$ locally compact?

Question

I think that it's locally compact but I don't truly know how to prove it. What I thought was that $[ 0,1)$ is closed and we have that a closed subset of a compact subset is compact so using that $[0,1]$ is compact and the product of compacts is compact I could prove that not only $[0,1) \times [0,1)$ was locally compact but it was compact.

IMPORTANT: My definition of locally compact is: every point $x$ of $X$ has a compact neighbourhood

Answer 1

Yes, the product of two locally compact spaces is locally compact: if $(x,y) \in X \times Y$ and $x$ has a compact neighbourhood $C_x$ and $y$ has a compact neighbourhood $C_y$ then $C_x \times C_y$ is a product neighbourhood of $(x,y)$ and is compact (as all products of compact sets are).

This generalises to finite products easily.

$[0,1)$ is not closed but $[0,1) \times [0,1)$ is open in $[0,1]\times [0,1]$ which is compact Hausdorff so your set is locally compact that way too.

Answer 2

If $(a,b) \in [0,1)\times [0,1)$ and $a>0$ and $b>0$ then $[a-r,a+r]\times [b-r,b+r]$ is a compact neighborhood of $(a,b)$ as long as $r>0$ is so small that $[a-r,a+r]\times [b-r,b+r] \subset [0,1)\times [0,1)$. If $a=0$ just change the left end point of $[a-r,a+r]$ to $0$ and if $b=0$ change the left end point of $[b-r,b+r]$ to $0$.