The usage of induction in the proof of Urysohn's lemma

Question

[Proof taken from]: Kelley's book "General Topology".

If $A$ and $B$ are disjoint closed subsets of a normal space $X$, then there is a continuous function $f$ on $X$ to the interval $[0,1]$ such that $f$ is zero on $A$ and one on $B$.

The proof is structured as follows (here $F(-)$ will be a family of sets in $X$):

1. Let $D$ the set of positive dyadic rational numbers. For $t$ in $D$ and $t > 1$ let $F(t) = X$, let $F(1) = X - B$ and let $F(0)$ be an open set containing $A$ such that cl$(F(0))$ is disjoint from $B$. (This part is clear)

2.For $t$ in $D$ and $0 < t < 1$ write $t$ in the form $t = (2m+1)2^{-n}$ and choose, inductively on $n$, $F(t)$ to be an open set containing cl$(F(2m2^{-n}))$ and such that cl$(F(t)) \subseteq F((2m+2)2^{-n}))$. This choice is possible because $X$ is normal.

Here there are my problems... How exactly does he use induction? And what exactly is the the role of $m$ during this operation? Please, is there someone that could explain me all the details regarding this point? I would to specify that:

• I found another related topic here (Urysohn's Lemma: Proof), but I actually don't understand more or less the fourth line of the proof (which is exactly my problem);

• I know that there are a lot of variants of this proof (for example the one on the book by Munkres, or another one on the classic by Rudin), but I strongly prefer to understand this one;

• I have searched on the web, but without finding any detailed explanation;

3. Let $f(x) = \inf\{t : x \in F(t) \}$. The previous lemma shows that $f$ is continuous. Furthermore, the function is zero on $A$ and one on $B$ [some related considerations]. (This part is clear).

Thank you very much for any hint/help/advice! Cheers

Answer 1

The induction goes as follows.

Step 1:
Define $F(1/2)$ using $F(0)$ and $F(1)$.

Step 2:
Define $F(1/4)$ using $F(0)$ and $F(1/2)$.
Define $F(3/4)$ using $F(1/2)$ and $F(1)$.

Step 3:
Define $F(1/8)$ using $F(0)$ and $F(1/4)$.
Define $F(3/8)$ using $F(1/4)$ and $F(1/2)$.
Define $F(5/8)$ using $F(1/2)$ and $F(3/4)$.
Define $F(7/8)$ using $F(3/4)$ and $F(1)$.

etc. (I assume the pattern is clear.)

At each step the definition of $F$ depends on values of $F$ defined in previous steps, and sometimes the definitions of $F(0)$ and $F(1)$ already made.

In the proof, we're defining $F(t)$ where $t=(2m+1)2^{-n}$. Note the $2m+1$ ensures the numerator of the fraction $\frac{2m+1}{2^n}$ is odd, so that $t$ can't be written with a lower power of $2$ in the denominator. (Example: if we're in Step 3, $n=3$ and $m=0,1,2$, or $3$.)

Given this, the values of $F$ at $(2m)2^{-n}$ and $(2m+2)2^{-n}$ have already been defined in previous steps, because both of these can be written with smaller powers of $2$ in the denominator (at most $2^{n-1}$ in the denominator, but sometimes even smaller, for example, in the $n=3$, $m=2$ case, $(2m)2^{-n}=\frac48=\frac12$).

So there is no problem with the induction.

Answer 2

Kelley is defining $F(t)$ for $t=(2m+1)2^{-n}$ in terms of two "previous" $F(t)$ involving $n-1$, namely $t=2m2^{-n}=m2^{-(n-1)}$ and $t=(2m+2)2^{-n}=(m+1)2^{-(n-1)}$.

It helps to write out the procedure for $n=1$, then $n=2$, and so on. So having defined $F(0)$ and $F(1)$, you then define $F(1/2)$ (i.e., $n=1$, and $m=0$), then you define $F(1/4)$ and $F(3/4)$ (i.e., $n=2$, $m=0, 1$), and so on.

The value of $m\in \{0, 1, \ldots,2^{n-1}-1\}$ is there to specify which dyadic $t$ you are working on. In the induction step $n-1\to n$, the value $m$ is generic.