This is related to 6.3.24 in Engelking's Topology book.

The Hilbert space $H$ is the set of sequences $(x_i) \in \mathbb R ^\omega$ such that $\|x\|=\sum _{i=1} ^\infty x_i ^2<\infty$, with metric $d(x,y)=\sqrt{\sum_{i=1} ^\infty (x_i-y_i)^2}$.

Let $X$ be the subspace of $H$ consisting of the sequences of rationals.

Observe that by assigning a point $(x_i)\in X$ with the same point in $\mathbb Q ^\omega$ one defines a continuous mapping and deduce that there is a one-to-one continuous mapping $f:X\to C$ of $X$ into the Cantor set $C$ such that $f(x_0)=0$.

By the observation we may assume $X$ is second countable and zero dimensional. Let $x_0\in X$. Let $\{U_i:i\in\omega\}$ be a clopen base for $X$. Let $A=\{i\in\omega:x_0\in U_i\}$ and $B=\omega\setminus A$. For each $x\in X$ let

$$f(x)(i)=\begin{cases} 0 &\mbox{if } (i\in A \wedge x\in U_i) \vee (i\in B \wedge x\in X\setminus U_i) \\ 1 & \mbox{else} \end{cases}.$$

Then $f:X\to 2^\omega$ by $f(x)=(f(x)(i))$ is well-defined, one-to-one, continuous, and $f(x_0)=0$.

Define a mapping $g:X\to [0,1]^2$ by letting $g(x)= \big( \frac{f(x)}{\max (1,\|x\|)},\frac{\|x\|}{1+\|x\|} \big)$. Show that the space $X_1=g[X]\cup \{(0,1/2)\}$ is hereditarily disconnected but not totally disconnected. Hint: Note that there is a one-to-one continuous mapping of $g[C]$ onto $f[X]$, and that if $A\subseteq X$ is clopen and $\{\|x\|:x\in A\}$ is bounded then $A=\varnothing$.

I assume that $g[C]$ should be $g[X]$.

$X_1$ is hereditarily disconnected: For $(r,s)\in g[X]$ let

$$h(r,s)=\begin{cases} r &\mbox{if } s\leq 1/2 \\ {\frac{rs}{1-s}} & \mbox{else} \end{cases}.$$

It is easy to show that $h$ is a continuous one-to-one mapping of $g[X]$ onto $f[X]$. As $f[X]$ is hereditarily disconnected, so is $g[X]$. Also, in creating this mapping I realized we can think of $g[X]$ as a subset of $C\times [0,1]$, squeezed at the top to the point $(0,1)$. It is now clear that no connected subset of $X_1$ meets $\{(0,1/2)\}$ and $g[X]$, as $(0,0)$ is the only point of $g[X]$ along the $y$-axis.

$X_1$ is not totally disconnected: Suppose $A$ is clopen in $X_1$ containing $(0,1/2)$, missing $(0,0)$. There exist $\epsilon,\delta>0$ such that $\big(B_\epsilon (0)\times B_\delta (1/2)\big)\cap X_1\subseteq A$. There exists $r\in (0,\epsilon)$ such that $[0,r]\cap C$ is clopen in $C$. Then I BELIEVE $$\big([0,r]\times [0,1/2]\big)\cap g[X]\setminus A$$ is clopen in $g[X]$. It is also nonempty as it contains $(0,0)$. Its preimage under $g$ is nonempty and clopen in $X$, and has bounded norm. This contradicts the claim in the 2nd part of the hint, which is provable via the standard argument that $X$ (the Erdos space) is not zero dimensional.



NOTE1: I believe I have solved this problem now, but I would of course appreciate any feedback/verification. Also, if anyone is interested I could add some details to my solution.

NOTE2: For the purpose of giving a space which is hereditarily disconnected but not totally disconnected, I think we could have defined $g$ simply by $g(x)=(f(x),\|x\|)$, and let $X_1=g[X]\cup \{(0,r)\}$ for any $r\in (0,\infty)$. The definition above is more complicated in order to achieve the squeezing, so that adding $(0,1)$ to the space actually makes it connected (by the 2nd part of the hint). That is the subject of another part of the exercise.

  • $\begingroup$ For the 1st part, you should let $ B=\{U_n\}_{n\in N}\cup \{U_n^c\}_{n\in N}$ be a base of clopen sets for $X$ with $ i\ne j\to U_i\ne U_j\ne U_i^c$ and with $ 0\in U_i^c$ and let $f(x)=\sum_{n\in N}2f_i(x)3^{-n}$ where $f_i(x)=1$ if $x\in U_i$ and $f_i(x)=0$ if $x\in U_i^c.$ I'm working on the second part. And you actually found a typo in Engelking! (I checked my copy.) $\endgroup$ Dec 25, 2015 at 2:38
  • $\begingroup$ I see that I need to change that part of my solution. How does it look now? By the way, I took for granted that $C\simeq 2^\omega$, so I left out the summation. $\endgroup$ Dec 25, 2015 at 3:07
  • $\begingroup$ @ForeverMozart: For the first part you should be working with a countable clopen base for $\Bbb Q^\omega$ and then composing with the map from $X$ to $\Bbb Q^\omega$. I can vouch for the typo that you found, since it’s been corrected in my edition. It would be helpful if you pointed out that $h$ is defined so that $h\circ g=f$. Finally, it’s true that a clopen nbhd $A$ of the extra point in $X_1$ contains points of $g[X]$ with arbitrarily small $x$-coordinates, and perhaps I’m being dense, but it’s not clear to me why this should ensure that $x_0\in g^{-1}[A]$. $\endgroup$ Dec 25, 2015 at 3:19
  • $\begingroup$ @BrianM.Scott Yes you are right about the first two things. Regarding your final point, you can pick a sequence of points $f(x_i)$ ($x_i\in g^{-1}[A]$ and $i\geq 1$) with $f(x_i)\to 0=f(x_0)$. Since we are dealing with metric spaces this implies $x_i\to x_0$. Since $g^{-1} [A]$ is closed it contains $x_0$. $\endgroup$ Dec 25, 2015 at 3:25
  • $\begingroup$ @ForeverMozart: I understood that last step. It’s getting a sequence in $g^{-1}[A]$ such that $f(x_k)\to 0$ that’s bothering me. How do you know that there are points of $X$ with norms close to $1$ that $f$ maps arbitrarily close to $0$? $\endgroup$ Dec 25, 2015 at 3:30


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