Infinite Concatenation of Homotopies

Question

In Chapter 0 of Hatcher's Algebraic Topology book, it is proven that CW pairs $(X,A)$ have the homotopy extension property (pg 15- I would include an image, but I don't have enough reputation to do that).

What I don't understand is the "infinite concatenation of homotopies" part. If we have deformation retractions $F_n:A_n\times [2^{-n-1},2^{-n}]\to B_n$, $A_n\subseteq A_{n+1}$,$B_n\subseteq B_{n+1}$ then how does one get a deformation retraction $F:\cup_nA_n\times I\to \cup_nB_n$ (assuming that the union has the weak topology induced by the inclusions of the $A_n$). The problem is that $\cup_n(A_n\times [2^{-n-1},2^n])$ does not cover $\cup_nA_n\times I$, so we can't glue the $F_n$s into $F$. I thought $F(x,t)$ should be defined by $F_n(x,t)$ where $n$ is the smallest integer for which $x\in A_n$, but then one does not need to take the $F_n$s on the intervals $[2^{-n-1},2^{-n}]$

Answer 1

First notice that $X^n\times I\subset X^{n+1}\times\{0\}\cup\left(X^n\cup A^{n+1}\right)\times I$, that is, in your notation $B_n\subset A_n\subset B_{n+1}$ (so $\cup A_n=\cup B_n=X\times I$).
Denote now the contracting homotopy of $X^n\times I$ to $X^n\times\{0\}\cup\left(X^{n-1}\cup A^n\right)\times I$, parameterized by $[1/2^{n+1},1/2^n]$, by $H_n$. extend it to $X^n\times I\cup A\times I\cup X\times\{0\}$ by identity (it makes sense since $H_n$ is the identity homotopy on $A^n\times I\cup X^n\times\{0\}$).
Now let $x$ be a point in $X^n\times I-A\times I$. What happens to x is that up to $t=1/2^{n+1}$, the point doesn't move, then in the interval $[1/2^{n+1},1/2^n]$ it moves according to $H^n$. At $t=1/2^n$, $x$ is either in $X\times\{0\}$ or in $A\times I$, and it stays there, or it ends in $X^{n-1}\times I$ and continues to move with the next homotopies. Formally, $$H(x,t)= \begin{cases} x & t<1/2^{n+1} \\ H_n(x,t) & t\in[1/2^{n+1},1/2^n] \\ H_m(H(x,1/2^{m+1}),t) & t\in[1/2^{m+1},1/2^m],m<n \end{cases}$$ this definition is recursive, but finite, and covers all of $X\times I$ because of the extension we made above.