show that there is a continuous function from [0, 1] onto a countable product of copies of [0, 1] with product topology

I only have difficulties in problem 11.

My efforts:

Follow the hint. We have a sequence of functions {$$F_k(t)$$}. Each $$F_k(t)$$ maps $$t\in$$ [0, 1] to a sequence $$(t_1,t_2,...)$$ in $$\Pi_{n\geq 1}[0,1]_n$$. In problem 10, we have constructed a continuous function $$f^{(k)}(t)$$ from [0, 1] onto $$[0, 1]^k$$. Now we make reference to a result from a previously solved problem.

Let $$X=[0,1]^k$$ and $$Y=\Pi_{n>k}[0,1]_n$$. Let $$y_0$$ be the all-zero sequence in $$Y$$. Then the map $$f:X\rightarrow X\times Y$$ defined by $$f(x)=x\times y_0$$ is continuous. Then $$F_k(t)$$ is the composite of two continuous functions: $$F_k(t)=f(f^{(k)}(t))$$. This holds for each fixed $$k$$. Thus each $$F_k(t)$$ is continuous. Of course it is not onto due to its 0-tail.

Recall that $$[0, 1]^k$$ is using the square metric on $$\mathbb{R}^k$$: $$d_k(x,y)=$$ max{$$|x_1-y_1|,|x_2-y_2|,...,|x_k-y_k|$$}. Also recall that $$\mathcal{C}(I,I^k)$$ is using the corresponding sup metric: $$\rho_k(f,g)=$$ sup{$$d_k(f(t),g(t))|t\in I$$}.

Now $$\Pi_{n\geq 1}[0,1]_n$$ is using product topology. We cannot treat it as a metric space. Accordingly there is no metric for $$F_k(t)$$. We cannot write something like $$\rho(F_7,F_9)$$. We cannot prove the convergence of {$$F_k(t)$$} by saying it is Cauchy. We can only use the open-neighborhood definition of convergence for {$$F_k(t)$$}. What's the open neighborhood of $$F_k(t)$$? We can view $$F_k(t)$$ as a point in the product space $$[0,1]\times\Pi_{n\geq 1}[0,1]_n=\Pi_{n\geq 0}[0,1]_n$$. Then the convergence of {$$F_k(t)$$} to $$F(t)$$ is defined as: Any open neighborhood of $$F(t)$$ intersects {$$F_k(t)$$}. More exactly, it is defined as: Given any $$n$$ and any $$\epsilon>0$$, {($$x,y$$)|max{$$|x-t|,d_n(y_{1:n},F(t)_{1:n})$$}<$$\epsilon$$} intersects {$$F_k(t)$$}. That is, given any $$n$$ and any $$\epsilon>0$$, there exists $$k$$ such that $$|t-s|<\epsilon$$ and $$d_n(F_k(s),F(t))<\epsilon$$. The convergence cannot be guaranteed by the construction of $$f^{(k)}(t)$$. We can find a $$F_k(s)$$ having the same 1-to-$$n$$ coordinates as F(t), but $$|t-s|$$ may be greater than $$\epsilon$$.

------------------------Updated according to the hints given by Brian M. Scott------------------------

Let $$h^{(k)}(t)$$ be the $$k$$-time composition of $$h(t)$$, e.g., $$h^{(3)}(t)=h(h(h(t)))$$.

$$f^{(2)}(t)=(g(t),h(t))$$

$$f^{(3)}(t)=(g(t),g(h(t)),h(h(t)))$$

$$f^{(4)}(t)=(g(t),g(h(t)),g(h^{(2)}(t)),h^{3}(t))$$

$$f^{(5)}(t)=(g(t),g(h(t)),g(h^{(2)}(t)),g(h^{(3)}(t)),h^{4}(t))$$

Given $$n$$, $$t_n^{(k)}=0$$ for $$k. We claim that $$t_n^{(k)}=g(h^{(n-1)}(t))$$ for $$k>n$$. We show this by induction on $$n$$.

By construction, $$f^{(k+1)}(t)=(g(t),f^{(k)}(h(t)))=(g(t),g(h(t)),f^{(k-1)}(h^2(t)))=(g(t),g(h(t)),g(h^{(2)}(t)),f^{(k-2)}(h^3(t)))=...$$

The $$n=1$$ case is that $$t_1^{(k)}=g(t)$$. This is obvious from the construction.

Assume it is true for all $$n$$. By construction, $$t^{(k)}_{n+1}=h(t)^{(k)}_n=g(h^{(n-1)}(h(t)))=g(h^{(n)}(t))$$.

We have shown that $$t_n^{(k)}$$ is constant after $$k>n$$.

Next we show $$F$$ is continuous. We only need to show: Given any open neighborhood $$U$$ of $$\langle t_n:n\geq 1\rangle$$, $$F^{-1}(U)$$ is open.

Since $$\prod_{n\geq 1}[0,1]_n$$ is using product topology, we can take $$U$$ to be $$U=\prod^N_{n=1}\big((t_n-\epsilon_n,t_n+\epsilon_n)\cap [0,1]_n\big)\times\prod_{n\geq N+1}[0,1]_n$$ for any fixed $$N$$. Since $$f^{(N)}$$ is continuous when $$[0,1]^N$$ is using square metric topology and the ball $$B(\langle t_n,n=1,...,N\rangle,\min\{\epsilon_1,\ldots,\epsilon_N\}, d_N)\subset\prod^N_{n=1}\big((t_n-\epsilon_n,t_n+\epsilon_n)\cap [0,1]_n\big)\;,$$ $$f^{(N)}$$ is also continuous when $$[0,1]^N$$ is using box topology. Thus $$F^{-1}(U)=(f^{(N)})^{-1}\left(\prod^N_{n=1}\big((t_n-\epsilon_n,t_n+\epsilon_n)\cap [0,1]_n\big)\right)$$ is open.

Finally we need to show $$F$$ is surjective: Given any $$\langle t_n,n\geq 1\rangle\in\prod_{n\geq 1}[0,1]_n$$, there exists a $$t\in[0,1]$$ such that $$F(t)=\langle t_n,n\geq 1\rangle$$. For any fixed $$N$$, we can find a $$t$$ such that $$g(h^{(n-1)}(t))=t_n$$ for all $$n by construction of $$F(t)$$. Since $$t$$ and $$t_n$$ remain constant as we increase $$N$$, we have found the desired $$t$$.

The product topology on $$\prod_{n\ge 1}[0,1]_n$$ actually is metrizable: this is true of any product of countably many metrizable spaces. However, you don’t need that here.

HINT: For $$t\in[0,1]$$ and $$k\ge 1$$ let $$F_k(t)=\left\langle t_n^{(k)}:n\ge 1\right\rangle$$.

• Show that for each $$n\ge 1$$ the sequence $$\left\langle t_n^{(k)}:k\ge 1\right\rangle$$ is eventually constant and therefore converges to some $$t_n\in[0,1]$$. (In fact you should find that $$t_1=g(t)$$, $$t_2=g(h(t))$$, $$t_3=g(h(h(t)))$$, and so on.)

Define $$F:[0,1]\to\prod_{n\ge 1}[0,1]_n:t\mapsto\langle t_n:n\ge 1\rangle$$.

• Show that $$F$$ is continuous.
• Show that $$F$$ is a surjection.
• Please see the last paragraph of my updated question description. Is my proof of surjection correct? – Junk Warrior Jul 11 '20 at 22:34
• @JunkWarrior: Yes, that looks okay. I fixed up some of the MathJax to make it easier to read. To get real angle brackets you can use \langle ($\langle$) and \rangle ($\rangle$); the max and min operators are \max ($\max$) and \min ($\min$); and the product symbol is \prod ($\prod$), just as the summation symbol is \sum ($\sum$). – Brian M. Scott Jul 11 '20 at 23:35

Another route to the same fact: we consider the Cantor set $$\mathbf{C} \subseteq [0,1]$$. This is well known to be homeomorphic to $$C=\{0,1\}^{\Bbb N}$$, the countable Cantor cube. The well-known Cantor function maps $$C$$ onto $$[0,1]$$ (sending a sequence $$(x_n)_n \in C$$ to $$\sum_n \frac{x_n}{2^{n+1}}$$). It's clear that $$C^{\Bbb N} \simeq C$$ (a reshuffle of the countably many coordinates on both sides) and so combining the mentioned homeomorphisms with the countable power of the Cantor function, we get a map $$\mathbf{C} \to [0,1]^{\Bbb N}$$ and we extend this by Tietze'e extension theorem to a surjective continuous map $$[0,1] \to [0,1]^{\Bbb N}$$.

Bonus fact: any compact connected and locally connected metric space is the continuous image of $$[0,1]$$ (one of the main theorems on so-called Peano continua). And $$[0,1]^{\Bbb N}$$ is metrisable as any countable product of metric spaces is (e.g. see here), and standard theorems tell us it's compact, connected and locally connected.