Let $X_1$, $X_2$, and $X_3$ be spaces.
(a) Prove that $(X_1 \times X_2) \times X_3$ is homeomorphic to $(X_1 \times X_2) \times X_3$ is homeomorphic to $X_1 \times (X_2 \times X_3)$
So, I think I have the idea behind this one, but was hoping for some second opinions, critiques, or fixes.
Suppose that $X_1$, $X_2$, and $X_3$ are spaces. Let $h: (X_1 \times X_2) \times X_3 \rightarrow X_1 \times (X_2 \times X_3)$ be defined as follows. $$ h((x,y),z) = (x,(y,z)) $$ To start we show that this function is continuous. Let $O$ be and open set in $X_1 \times (X_2 \times X_3)$ Then $O = O_1 \times (O_2 \times O_3)$ Where $O_i$ is and open set in $X_i$. Since $O_i$ is open in each $X_i$ the set $O_* = (O_1 \times O_2) \times O_3$ must be an open set. As it is the Cartesian product of open sets. Since $O_* \subset (X_1 \times X_2) \times X_3$, we have $h$ continuous. In addition we see that $h$ is clearly 1-1. To show that the inverse is continuous we observe that we need to show that there exists are open sets $V,U$ in $(X_1 \times X_2) \times X_3 $ such that $h(V) \subset H(U)$. To show this let $U = (U_1 \times U_2) \times U_3$ where each $U_i$ is open in $X_i$, and let $V = (V_1 \times V_2) \times V_3$ where $V_i \in X_i$ is open and $V_i \subset U_i$. Then $h(V) = V_1 \times (V_2 \times V_3)$ and $h(U) = U_1 \times (U_2 \times U_3)$. Since each $V_i \subset U_i$ $h(V) \subset h(U)$ Therefore we have the two spaces are homeomorphic.
I think this is the correct argument, but it seems a little redundant. I will likely uses a similar proof to show that
(b) $X_1 \times X_2$ is homeomorphic to $X_2 \times X_1$, using $h(x,y) = (y,x)$, but wanted to get some feed back to make sure I have the right Idea first.
Any help and critiques are very appreciated.