This question is based off a problem from Classical Descriptive Set Theory by Kechris. In this book, Kechris makes the claim that, when $\{0, 1\}$ is endowed with the discrete topology, then the Cantor space $\{0, 1\}^{\mathbb{N}}$ is homeomorphic to the set $\{0, 1\}^{\mathbb{N}^{<\mathbb{N}}}$ (both endowed with the product topology). I am struggling to see why this is true, and I cannot think of an explicit homeomorphism between the two spaces. Could someone explain why this is the case?
Why is $2^{\mathbb{N}}$ Homeomorphic to $2^{\mathbb{N}^{< \mathbb{N}}}$
cantor setdescriptive-set-theorygeneral-topologypolish-spacesproduct-space
Related Solutions
Your space $C$ is the collection of functions $f:\omega=\{0,1,2,\ldots\}\to \{0,1\}=X$, i.e. the collection of infinite binary sequences. I will try to illustrate how we can understand the space $C$ and its topology below, deriving some of its properties.
The basic open sets in this collection are (by definition) infinite products of open sets in $X$ where all but finitely many factors are $X$. Suppose the factors correspond to a sequence $k_1<k_2<\cdots<k_s$ of natural numbers. At these factors (assuming the set is not empty) there correspond either the open set $\{0\}$ or the open set $\{1\}$. So we can associate to each basic open set $O$ a function from a finite subset $F=\{k_1,\ldots,k_s\}$ of $\omega$ to $X$. This is a bijection. Note that if $f,g\in O$ and $F$ is associated to $O$; then $f\mid F=g\mid F$.
Now take a function $f:\omega\to X$. We can consider the open sets $O_i=\prod_{k=0}^i \{f(k)\}\times X\times \cdots$. Note that given any basic open set $O$ that contains $f$, there is $j$ such that $O_j\subseteq O$ (can you prove this?). This means that the collection $\mathscr B=\{O_i:i\in\omega\}$ is a neighborhood basis for $f$. This means our space is second countable! It is also separable, since the collection of functions that are eventually constant is countable and dense: given a basic neighborhood $O_i$ of $f$, the function that agrees with $f$ on $\{0,1,2,\ldots,i\}$ and is constant afterwards is in $O_i$. This proves this space is also perfect.
In particular $g\in \overline{\{f\}}$ iff $g$ agrees with $f$ on $\{0,1,\ldots,i\}$ with each $i$. This means $g=f$; so singletons are closed $\{f\}$. In fact your space is Hausdorff: if $g$ and $f$ do not agree at position $k$ (say one takes the value $0$ and the other $1$), then $\cdots\times X\times \{0\}\times X\times\cdots$ and $\cdots\times X\times \{1\}\times X\times\cdots$ are disjoint open sets that contain them.
Actually, your space is metrizable, and the neigborhood basis we obtained gives a decent idea of what this metric is: we define $d(f,g)=2^{-\nu(f,g)}$ where $\nu(f,g)$ is the first $k\in\omega$ such that $f(k)\neq g(k)$. We define $\nu(f,f)=\infty$, of course! Note then that $O_i=\{g\in C:d(f,g)<2^{-i}\}$. Check that $d(f,g)$ is indeed a metric.
Finally, your space is compact. Since we've shown it is metrizable, we can simply check it is sequentially compact. Now take a sequence of functions $(f_i)$ in your space. If we look at $f_i(0)$ for $i=1,2,3,\ldots$, we get a sequence of zeros and ones. Since this sequence is infinite, either the ones or the zeros repeat infinitely often. So we can take a subsequence indexed by $N_0$ that has all its first coordinates equal to say $x_0\in \{0,1\}$. Now look at the second coordinate of this sequence, and repeat the process to get $N_1\subseteq N_0$ and some $x_1$. We get a sequence of nested infinite subsets $N_0\supseteq N_1\supseteq N_2\supseteq \cdots$, and we may thus take a subsequence $f_{k_j}$ such that $k_i\in N_i-N_{i+1}$. You can check that this subsequence converges to the $f$ such that $f(i)=x_i$ (note this is just another diagonalization trick!).
Given we have exhibited your space as a metric space, I believe you can understand its topology quite nicely. Now, note that the Cantor set consists of real numbers whose ternary expansion contains only $0$s and $2$s. Does this give you an idea of what a possible homeomorphism $\eta:C'\to C$ may be?
For 2. $K(X)$ is Polish, the intersection of countably many dense open sets is dense and hence nonempty, so there are plenty of $K$ on which $f$ is injective.
For 1. I would try something like this: take a countable base, $\mathcal{B}$, for the topology of $X$. For a pair $\langle B, C\rangle$ of elements of $\mathcal{B}$ with disjoint closures let $$U(B,C)=\{K \in K(X): f[K\cap B]\cap f[K\cap C]=\emptyset\}\text{.}$$
Then prove that $U(B,C)$ is open and dense.
Best Answer
Let $I.J$ be infinite index sets. Then $\{0,1\}^I \simeq \{0,1\}^J$ (as spaces) iff the sets $I$ and $J$ have the same cardinality. A shuffle of coordinates is a homeomorphism (if $f: I \to J$ is a bijection, $\hat{f}: \{0,1\}^J \to \{0,1\}^I$ defined by $\hat{f}(g)=g \circ f$ is a homeomorphism)
For the reverse note that the weight of $\{0,1\}^I$ equals $|I|$ and weight ( the minimal size of a base) is a topological invariant.