Lemma concluded by Urysohn's lemma, continuity, proof explanation

Question

Lemma 2: Let $X$ be a $T_4$ space. Let $A\subseteq X$ closed and $\varphi: A\to [-1,1]$ continuous. Then exists $\Phi: X\to [-1,1]$ continuous with $\Phi_{|A}=\varphi$.

I have a problem with the given proof to this statement. This lemma uses the following lemma, concluded by Urysohn's lemma:

Lemma 1: Let $X$ be a $T_4$ space. Let $A\subseteq X$ closed and $\varphi: A\to [-c,c]$ continuous, $c>0$. Then exists $\psi: X\to [-\frac{c}{3},\frac{c}{3}]$ continuous with $|\varphi(a)-\psi(a)|\leq\frac23c$ for every $a\in A$.

I have a problem understanding the part in the proof of Lemma 2 where it is shown that the constructed function $\Phi$ is continuous.

Proof:

By lemma 1 it existts $\psi_0: X\to [-\frac13,\frac13]$ continuous with $|\varphi(a)-\psi_0(a)|\leq \frac23$ for every $a\in A$.

Repeated use of lemma 1 gives $\psi_1: X\to\mathbb{R}$ continuous with $|\psi_1(p)|\leq\frac13\cdot\frac23$, $|\varphi(a)-\psi_0(a)-\psi_1(a)|\leq\frac23\cdot\frac23$.

Then $\psi_n:X\to\mathbb{R}$, $|\psi_n(p)|\leq\frac13\cdot\left(\frac23\right)^n$ and

$|\varphi(a)-\psi_0(a)-\dotso -\psi_n(a)|\leq \left(\frac23\right)^n\cdot\frac23$

Let $\Phi(p)=\sum_{n=0}^\infty \psi_n(p)$ then $|\Phi(p)|\leq \frac13\sum_{n=0}^\infty |\psi_n(p)|=1$ so $\Phi: X\to [-1,1]$.

For $a\in A$ is $\Phi(a)=\varphi(a)$.

So far so good. Now the part, where I do not really understand what is going on...

Why is $\Phi$ continuous?

Let $p\in X$, $\varepsilon >0$. Then exists $m\geq 0$ such that $\frac13\sum_{n=m+1}^\infty \left(\frac23\right)^n\leq\dfrac\varepsilon3$.

Let $V$ be a neighborhood of $p$ such that for every $q\in V$ holds:

$\color{red}{|\sum_{n=0}^m (\psi_n(p)-\psi_n(p))|\leq\dfrac\varepsilon3}$.

This yields $\color{red}{|\Phi(p)-\Phi(p)|\leq\dfrac\varepsilon3+\dfrac\varepsilon3+\dfrac\varepsilon3=\varepsilon}$ for $q\in V$.

I am reading hand written lecture notes, so this might be due to bad handwriting, but I am pretty sure in the notes it reads exactly like this, what does not make sense at all, does it? There is not even $q$ involved at all here.

So one $p$ in every inequality is for sure supposed to be $q$.

How do we derive the last inequality then?

$|\Phi(p)-\Phi(q)|=|\sum_{n=0}^\infty (\psi_n(p)-\psi_n(q))|$

This should be done like this:

$=|\sum_{n=0}^\infty (\psi_n(p)-\psi_n(q))|=|\sum_{n=0}^m (\psi_n(p)-\psi_n(q))+\sum_{n=m+1}^\infty (\psi_n(p)-\psi_n(q))|$

$\leq |\sum_{n=0}^m (\psi_n(p)-\psi_n(q))+\sum_{n=m+1}^\infty |\psi_n(p)|+\sum_{n=m+1}^\infty |\psi_n(p)|~|\leq \dfrac\varepsilon3+\dfrac\varepsilon3+\dfrac\varepsilon3=\varepsilon$

Can you confirm/explain this?

Thanks in advance.

Answer 1

Yes, essentially it's the following idea:

Suppose $f_n : X \to \Bbb R$ are continuous functions on $X$ with $\forall x \in X: |f_n(x)| \le a_n$ for each $n$ and $a_n \ge 0$ and such that $\sum_n a_n < \infty$. Then $f(x)=\sum_n f_n(x)$ is continuous.

To show continuity at $p$, take $\varepsilon>0$ and take $N$ such that $\sum_{n > N} a_n < \frac{\varepsilon}{3}$ and then also use that (this is missing from your argument I believe) that $f_N:= \sum_{n=1}^N f_n(x)$ is continuous as a finite sum of continuous functions, so we have a neighbourhood $U_p$ of $p$ such that $$|f_N(p)-f_N(x)| < \frac{\varepsilon}{3}$$ for $x \in U_p$, Now neutralise the tails indeed:

$$|f(p)-f(x)| = \left| (f_N(p)-f_N(x)) + \sum_{n > N} f_n(p) + \sum_{n > N} -f_n(x)\right| \le \\|f_N(p)-f_N(x)| + \sum_{n > N} |f_n(p)| + \sum_{n > N} |-f_n(x)| \le |f_N(p)-f_N(x)| + \sum_{n > N} a_n + \sum_{n > N} a_n < 3\frac{\varepsilon}{3} = \varepsilon $$

for all $x \in U_p$, as required.

So also use the continuity of the finite sum.