I was wondering if somebody can help me with completing a proof of the claim made above. The problem I am having occurs at the end of the proof, so I don't want to bring it up over here without context. I would like to add that this is not homework for any class. I'm studying topology on my own for fun.
I apologise if you find that proving this claim instead of for arbitary cartesian products of path connected spaces is particularly reductive. I intend to prove that next. I just wanted to prove this as a lemma first in order to get a feel for the proof of the larger claim. Also, please let me know if you need to know any definitions.
With all of that out of the way, here is what I have thus far:
Recall: A space X is said to be path connected if every pair of points of X can be joined by a path in X.
Let X and Y be path connected topological spaces. Suppose they are non-empty, for the result is trivial otherwise. Let $x_1$ and $x_2$ be elements of X and let $y_1$ and $y_2$ be elements of Y. We are interested in showing that there is a path from $x_1$ x $y_1$ to $x_2$ x $y_2$. There exists a continuous map $f: [a,b] \rightarrow X$ of some closed interval of $\mathbb{R}$ into X, such that $f(a) = x_1$ and $f(b) = x_2$ because X is path connected. There also exists a continuous map $g: [c,d] \rightarrow Y$ of some closed interval of $\mathbb{R}$ into Y, such that $f(c) = y_1$ and $f(d) = y_2$ because Y is path connected.
Let $A = [a,b]$ x $[c,d]$. Define $f': A \rightarrow X$ by $f'(x) = f(\pi_1(x)) = f \circ \pi_1 (x)$ and $g': A \rightarrow Y$ by $g'(x) = g(\pi_2(x)) = g \circ \pi_2 (x)$.
Are these projections onto the first and second factors ($\pi_1$ and $\pi_2$, respectively) continuous? We'll start with $\pi_1: $ X x Y $ \rightarrow X$, where X and Y are arbitrary topological spaces. Let U be open in X. We want to show that $\pi_1 ^{-1}(U)$ is open in X x Y. Notice that $\pi _ 1$ need not be injective, but it is certainly surjective. Notice that U x Y $\subset \pi _ 1^{-1}(U)$ is open in X x Y. Is it true that $\pi _1^{-1}(U) \subset$ U x Y? Fix u x v $\in \pi _ 1^{-1}(U)$. It is clear that $u \in U$, for if it wasn't, then $\pi _1 ($u x v$) \notin U$ (which cannot be). That $v \in Y$ is trivial. Hence, u x v $\in$ U x Y and $\pi _ 1^{-1}(U) \subset$ U x Y. Hence, $\pi _ 1^{-1}(U) = $ U x Y. We conclude that $\pi _ 1$ is continuous. Similarly, $\pi _2$ is continuous.
Knowing that, since $f$ and $g$ are continuous, we note that $f'$ and $g'$ are continuous by Theorem 18.2(c).
Let $h: A \rightarrow$ X x Y be given by the equation $h(a) = (f'(a),g'(a))$. We conclude that $h$ is continuous by Theorem 18.4 (Maps into products).
The problem I am having is whether or not [a,b] x [c,d] = [a x c, b x d]. This is needed to show that h is a path in X x Y.
This might be a particularly stupid thing to get caught up on, but I feel like I'm missing something here. I don't know what the order relation for the latter set is (I am aware, of course, that it is a subset of $\mathbb{R}$ x $\mathbb{R}$). Is it the dictionary order relation?
If anybody has any comments on the proof (mistakes, improvements, etc), please let me know. I'm fairly new to higher-level math (I'm a physics student) as I have only had a year of exposure to rigorous math, so any criticism is greatly appreciated. :) Also, thanks in advance for the help!
Theorems:
Theorem 18.2(c): Let X,Y, and Z be topological spaces. If $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ are continuous, then the map $g \circ f: X \rightarrow Z$ is continuous.
Theorem 18.4 (Maps into products): Let $f: A \rightarrow$ X x Y be given by the equation $$f(a) = (f_1(a),f_2(1)).$$ Then $f$ is continuous iff the functions $f_1: A \rightarrow X$ and $f_2: A \rightarrow Y$ are continuous.
Let $(x,y)$ and $(z,w)$ be two points in $X\times Y.$ Since $X$ is path-connected $X\times \{y\}$ is path connected (why?).So you have a path, say $f$ from $(x,y)$ to $(z,y)$. By a similar argument, you have a path $g$ from $(z,y)$ to $(z,w)$ since $\{z\}\times Y$ is path connected. Concatenating $f$ and $g$ you get the required path.