I want to prove that, given non-empty spaces $(X,\tau)$ and $(X',\tau')$, if
- $(X\times X',\tau\ast\tau')$ is connected, then $(X,\tau)$ is connected.
- $(X\times X',\tau\ast\tau')$ is compact, then $(X,\tau)$ is compact.
- $(X\times X',\tau\ast\tau')$ is Hausdorff, then $(X,\tau)$ is Hausdorff.
My attempt is to use the given (for the purposes of this class) claim that if $(X,\tau)$ is connected and $R\subseteq X\times X$ is an equivalence relation, then $(\frac{X}{R},\tau_q)$ is connected.(Same for connectedness) Thus if one defines an equivalence relation from $X\times X$ to $X$, the problem is solved. I am having trouble defining such a generic equivalence relation for this space, and further showing that $\tau_q = \tau$ necessarily. Thus I would appreciate some further explanation on the intuition of equivalence relations on topologies more generally.
I can, for example, prove this to be true for $(\mathbb{R}\times \mathbb{R}, \tau_{st} \ast \tau_{st})$ by defining the the relation $$(x,y)\sim (x',y')\quad if\quad x=x'$$Thus $[x]=\mathbb{R}$ and by the above $(X,\tau_{st})$ is connected. In this specific instance, for example, I have a hard time formalizing how the equivalence relation would affect(bad diction perhaps?) the topology. If the quotient topology is defined by $$\tau_{q}:=\{U\subseteq \frac{X}{R}\mid \pi^{-1}\in \tau \}$$ how would I demonstrate that $T_{st}$ is the topology generated from this?
Lastly, This is not a property for Hausdorff spaces if I am not mistaken. I would appreciate a counter example to disprove number 3.
Thanks in advance.
The first two are given by the fact that projections are continuous functions and the continuous image of a compact/connected space is compact/connected.
If $X$ is not Hausdorff, $X\times Y$ is not Hausdorff either: Fix $x_1$ and $x_2$ in $X$ that can’t be separated and any $y\in Y$, then $(x_1,y)$ and $(x_2,y)$ can’t be separated in $X\times Y$. Which gives 3.