I'm studying for an exam in topology; this is a question from a previous exam several years ago -- it's not being graded, I just want to know how to handle it. I'm more concerned here with learning the thought process involved in answering the following question(s) than I am with just being given the actual answers. If someone can provide a little guidance or maybe a hint or two, that would be great!

The question is this:

Consider the following result, which you can assume to be true without a proof:

(*) For every uncountable closed subset $F$ of $\mathbb{R}$ there exists a continuous function from $F$ onto the unit interval $I=[0,1]$.

Which of the following variations of (*) is true? Justify your answers.

(a) For every uncountable closed subset $F$ of $\mathbb{R}$ there exists a continuous function from $F$ onto $\mathbb{R}$.

(b) For every uncountable closed subset $F$ of $\mathbb{R}$ there exists a continuous function from $F$ onto the unit square $I^2=[0,1]^2$.

(c) For every uncountable closed subset $F$ of $\mathbb{R}$ there exists a continuous function from $F$ onto the Cantor ternary set $C$.

The work so far...

The definition of continuity that I'm working with is this: A function $f:X \rightarrow Y$ is continuous if for every set V open in Y, the set $f^{-1}(V)$ is open in $X$. This tells me that, in some sense, the topology on the domain must have more (or an equal number of) open sets than the topology on the range space.

I'm also aware that several properties are preserved under continuous maps: compactness, convergence of sequences, connectedness, path connectedness, the Lindelof property, and separability.

Statement (a) then should be false. $F$ is compact, as it is either a closed interval or a union of closed intervals in $\mathbb{R}$. The image of $F$ under any continuous function would also have to be compact. However, $\mathbb{R}$ is not compact. Thus such a continuous function does not exist.

I'm inclined to think that (b) is true, mainly because I can't come up with a reason why it wouldn't be. $I^2$ is not path connected, but $F$ isn't necessarily either. $I^2$ is connected and compact, but I don't think that really gives me anything to rule out the existence of the desired function. Now, if I want to show that such a function does exist, the only convincing argument I can come up with is to define one, but I'm not really sure how to do that here.

I think (c) is false, mostly based on intuition. I think it's going to come down to showing that the cardinality of the Cantor set is too big for a surjective function to exist, but I wouldn't put money on it.

Any help at all would be fantastic, but bonus points for putting a lot of detail into your thought process! I really want to learn how to handle this type of problem on my own, not just have an answer to this particular problem.

  • $\begingroup$ (a) is false, but your reasoning is not correct, for example you could take $F$ to be the reals in which case the identity is a suitable map. You can just produce a counterexample by taking $F$ to be compact. $\endgroup$
    – copper.hat
    Jul 30 '16 at 19:04

You are right about (a) (although you should be careful: a finite union of closed intervals is compact, but an infinite union need not be - consider $[0, 1]\cup [2, 3]\cup [4, 5]\cup . . . $). EDIT: Actually I misread it - your reasoning for (a) is wrong. How do you know $F$ is compact? Rather, the point is that there exists an uncountable closed set which is compact, and such a set is a counterexample to (a).

Your guess is right for part (b), but your reason is wrong (note that both $I^2$ and $I$ are path connected). HINT: it would be enough to find a continuous map from $I$ onto $I^2$ - do you see why, and have you heard of the Peano curve?

Your guess for (c) is right, but again, for the wrong reason: the Cantor set and $I$ have the same cardinality. Instead, consider (similarly to (a)): what can you say about the image of a connected set under a continuous map?

  • $\begingroup$ For (a) then, my logic should still suffice, right? In order for FF to be a closed subset of RR, it would have to be a finite union of closed intervals, so the rest of the argument should hold after I insert that condition that FF is a finite union of closed intervals For (b), I confused the unit square with the ordered square; thanks for pointing that out. I have heard of the Peano curve, but not in the context of topology (really, I only know that it's a fractal/space-filling curve). I'm not sure how I would define this function in topological terms. $\endgroup$
    – Keith
    Jul 30 '16 at 18:55
  • $\begingroup$ @Keith Your second sentence ("In order . . .") is not true: the set I describe above is closed but not compact, and not a finite union of closed intervals. But this doesn't effect your reasoning for (a): all you need is a single uncountable closed compact set, and $I$ works. Re: the Peano curve, you don't need to define it differently in topology - the point is just that it's a continuous surjective map from $I$ to $I^2$. $\endgroup$ Jul 30 '16 at 18:56
  • $\begingroup$ For (a), I see what you're saying. For (b), I'm not sure I see why it's enough to find a continuous map from $I$ onto $I^2$. Is it true in general that all closed subsets of $\mathbb{R}$ are homeomorphic with one another (and is the analogous statement true for arbitrary $\mathbb{R}^n$)? Or is there some deeper reason I don't see? $\endgroup$
    – Keith
    Jul 30 '16 at 23:12
  • $\begingroup$ @Keith Suppose I find a continuous map from $I$ onto $I^2$. The question allows you to assume already that there is a continuous map from any uncountable closed $F$ onto $I$. Do you see how to combine these? (And it is very false that any two closed subsets of $\mathbb{R}$ are homeomorphic - consider e.g. the closed sets $\emptyset$, $\{0\}$, $\{0, 1\}$, $[0, 1]$, $[0, 1]\cup [2, 3]$, $\mathbb{R}$.) $\endgroup$ Jul 30 '16 at 23:17
  • $\begingroup$ Also see my edit re: (a). $\endgroup$ Jul 30 '16 at 23:20

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