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There is a question on this here and an answer which is more like a hint. I am trying to fill in the details. I am having trouble with the last step.

Let $X$ be a compact metric space.

$1).$ Give $\left \{ 0,1 \right \}$ the discrete topology, and $2^{\mathbb N}$ the product topology, which is induced by the metric: $d(x,y)=1/n$ where $n$ is the least integer such that $x_n\neq y_n$.

$2).$ There are integers $1\le j_1\le m_1,\ $points $x_{j_1}$ and balls $B_{j_1}(x_{j_1})$ of diameter $1$, that cover $X$. Having chosen $B_{j_1,j_2,\cdots, j_{n-1}}(x_{j_1,j_2,\cdots, j_{n-1}})$ of diameter $1/(n-1)$, with $1\le j_k\le m_k,\ 1\le k\le _{n-1},$ we may choose $B_{j_1,j_2,\cdots, j_{n}}(x_{j_1,j_2,\cdots, j_{n}})$ of diameter $1/n$, with $1\le j_n\le m_n\ $. i.e. within each ball of diameter $1/(n-1)$ in the sequence, there is a finite collection of balls of diameter $1/n$. So the induction proceeds.

Thus, for each tuple $(j_1,j_2,\cdots, j_n,\cdots )$ there is a sequence of nested balls whose diameter $\to 0$, and whose intersection contains exactly one point. This gives a well-defined

$g:\prod^{\infty } _{k=1}\left \{ 1,2,\cdots,m_k \right \}\to X$. Clearly, $g$ is a surjection.

Now, for each $m_k$, choose the first $l_k$ such that $2^{l_k}\ge m_k$, take

$\prod^{\infty } _{k=1}\left \{ 1,2,\cdots,2^{l_k} \right \}$ and on each tuple for which there is a $k$ and a $q$ such that $m_k<q\le 2^{l_k},\ $define $f$ by evaluating $g$ on the tuple obtained by replacing $q$ with $m_k$.

$f$ is continuous since if $x$ and $y$ agree on the first $n$ coordinates, then they are both in a ball of diameter $1/n$.

$3).$ Define $h:2^{\mathbb N}\to \prod^{\infty } _{k=1}\left \{ 1,2,\cdots,2^{l_k} \right \}$ as follows:

Write each $x\in \left \{ 1,2,\cdots ,2^{l_k} \right \}$ as a sum $\sum_{i=1}^{l_k}\alpha^{(k)}_i2^{i}$ where each $\alpha^{(k)}_i$ is either $0$ or $1$.

Then, choose $\alpha= (\alpha^{(1)}_1,\cdots, \alpha^{(1)} _{2^{l_1}},\alpha^{(2)}_1,\cdots, \alpha^{(2)} _{2^{l_2}},\cdots )\in 2^{\mathbb N}$ and set $h(\alpha )=x$.

But how do we define $h$ on all of $ 2^{\mathbb N}?$

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Matematleta
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1 Answers1

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For $k\in\Bbb Z^+$ let $f_k$ be any surjection from $\{0,1\}^{\ell_k}$ to $\{1,\ldots,m_k\}$. Let $h:2^{\Bbb N}\to\prod_{k\in\Bbb Z^+}\{0,1\}^{\ell_k}$ be the obvious homeomorphism. (Just split each sequence in $2^{\Bbb N}$ into chunks of lengths $\ell_1,\ell_2,\ldots\;$.) Then $h\circ g:2^{\Bbb N}\to X$ is a continuous surjection.

Brian M. Scott
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