If $(X,\tau)$ is Hausdorff, is the product topology $X\times X$?
I have a feeling that some part of $\Bbb R$ to be Hausdorff, means there is isolation for all elements on the line, and if that is the case, then we get these isolated points when put onto $\Bbb R^2$ and they are still isolated in $\Bbb R^2$. This logic is really iffy to me, but that's my intuition.
Yes.
Let $p,q,r,s\in X$ s.t. $\langle p,q\rangle\neq\langle r,s\rangle$ so that $p\neq r\vee q\neq s$.
Let's say that $p\neq r$.
Since $X$ is Hausdorff open disjoint sets $U,V$ exist with $p\in U$ and $r\in V$. Then $\langle p,q\rangle\in U\times X$ and $\langle r,s\rangle\in V\times X$.
The sets $U\times X$ and $V\times X$ are disjoint and belong to the producttopology.
If $q\neq s$ then a practically similar proof can be given.
Yes. This is actually a special case: the product of arbitrarily many Hausdorff spaces is a Hausdorff space with the product topology. See a proof here.
If $X$ and $Y$ are Hausdorff topological spaces, then so is $X\times Y$ with the product topology.
Recall that product topology is generated by the sets of the form $U\times V$, where $U$ is open in $X$ and $V$ is open $Y$.
Now suppose that $(x_1,y_1)\ne(x_2,y_2)$ are two points in $X\times Y$. You want to find disjoint neighborhoods of these two points.
The fact that $(x_1,y_1)\ne(x_2,y_2)$ means that either $x_1\ne x_2$ or $y_1\ne y_2$.
If $x_1\ne x_2$ then there are two open subset $U_1\ni x_1$, $U_2\ni x_2$ such that $U_1\cap U_2=\emptyset$. (Here we are using the fact that $X$ is Hausdorff.)
Can you use $U_1$ and $U_2$ to find disjoint open subsets of $X\times Y$ which contain $(x_1,y_1)$ and $(x_2,y_2)$?
Can you see that the case $y_1\ne y_2$ is analogous?
