5
$\begingroup$

Let $C$ be the Cantor set, and $\omega$ the discrete space of integers $\{0,1,2,...\}$.

My conjecture:

(1) For each $n<\omega$ let $f_n:C\to C$ be a continuous function (possibly not onto). Let $$X=(\omega\times C)/\sim.$$ This is the quotient of $\omega\times C$ under the relation $\sim$: $$\langle n,c\rangle\sim\langle m,d\rangle\iff(\exists l\geq n,m)(f_l\circ ...\circ f_n(c)=f_l\circ ...\circ f_m(d)).$$ Then $\dim(X)=0$.

Is it true? I have not found it anywhere, despite doing a Google and looking in Dimension Theory texts.

The space $X$ is sometimes called a direct limit https://en.wikipedia.org/wiki/Direct_limit.

By $\dim(X)=0$, I mean $X$ has a basis of clopen sets (small inductive dimension is $0$).

More generally, I suppose one could ask if the direct limit of zero-dimensional spaces has dimension zero. But letting each factor be equal to the Cantor set potentially makes things easier.

What leads me to believe the statement is true, is two extreme cases of $f_n$'s. If each is constant, then $X$ is a singleton. And if each is the identity, $X$ is an increasing union of $\omega$-many Cantor sets. In this regard, the statement seems to be a generalization of the well-known "countable sum theorem" for (separable metric) spaces:

(2) If a separable metric space $X$ can be represented as the union of a sequence $F_0 , F_1, ...$ of closed zero-dimensional subspaces, then X is zero-dimensional.

I'm unsure if this is question is "research-level", so feel free to move to MSE if appropriate.

EDIT: Just to be clear, I am asking for either a reference or a proof of (1). In the meantime I'll try to prove it by mimicing the proof of the (2), which can be found on page 20 (item 1.3.1) of this book: http://www.maths.ed.ac.uk/~v1ranick/papers/engelking.pdf

Improve this question
$\endgroup$
1
  • $\begingroup$ Just a remark: Consider $C=A^{\omega}$, where $A$ is a finite set with $|A|\ge 2$, and $f_n=f$ the shift $(a_n)_{n\ge 0}\mapsto (a_{n+1})_{n\ge 0}$. Then the direct limit is the quotient of $C$ by the equivalence relation "to be eventually equal". The quotient topology is the indiscrete topology. This is not a counterexample (since it has an obvious basis of clopen subsets), but indicates that we should keep in mind that the direct limit can fail to be Hausdorff. $\endgroup$
    – YCor
    Commented May 4, 2018 at 9:19

1 Answer 1

Reset to default
4
$\begingroup$

It's not true. Here's an easy way to get $[0,1]$ as a limit.

Consider $C_0$ as the set of all sequences of $0,1$: each such sequence defines a binary expansion of some element in $[0,1]$. Let $C_n$ be obtained from $C$ by identifying, for each dyadic number of the form $k/2^n\in \mathopen] 0,1\mathclose[$ (possibly not reduced), its two dyadic expansions. Then $C_n$ is still homeomorphic to the Cantor space, and the inductive limit is naturally homeomorphic to $[0,1]$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ ah, very good. I thought of a counterexample in my car today, but yours is simpler. $\endgroup$
    – Forever Mozart
    Commented May 4, 2018 at 23:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .