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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.

proof:

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.

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  • $\begingroup$ It's incorrect to say that because $O$ is open it is a product of open sets of X1, X2, X3. However those sets form a basis for the topology so you only need to check h respects sets that are products. $\endgroup$ – spitespike Apr 15 '15 at 5:33
  • $\begingroup$ @ spitespike Sorry, I have not done much with the product topology. Can you elaborate a little on what it means for $h$ to respect the sets that are products. $\endgroup$ – Ben Apr 15 '15 at 5:39
  • $\begingroup$ The first part of your argument is great. If you rephrased it to say 1×(2×3) open iff (1×2)×3 is open you can dispense with the discussion of the inverse altogether as you have an open, continuous bijection $\endgroup$ – spitespike Apr 15 '15 at 5:40
  • $\begingroup$ Lots of open sets in the product topology aren't directly products of open sets. Think of the open unit ball in R^2. But any open set is the union of products of open sets, so it suffices to check that h^-1 of any product of open sets is open. That's all I mean by respects: that the inverse image will be open, which is exactly what you checked! $\endgroup$ – spitespike Apr 15 '15 at 5:46
  • $\begingroup$ So, my argument is okay if I refer to O as a basis element opposed to an open set? Edit disregard. I just found a theorem that Refers to this comment. $\endgroup$ – Ben Apr 15 '15 at 5:53

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