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There are two questions:

  1. Image of a countable set of real numbers under any continuous function is countable?

My claim is yes. Let $X$ is countable $\implies X=\{x_1,x_2,\ldots,\}$. Now $f(X)=\{f(x_1),f(x_2),\ldots,\}$ which can be atmost countable. Now my question is "What is the role of continuity here?"

  1. Image of a uncountable set of real numbers under any non-constant continuous function is uncountable?

I feel this is true. But unable to proceed. Please provide me a hint.

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  • $\begingroup$ If the functions are from the real line to the real line then mean value theorem gives the complete answer. Just go through what is mean value theorem. $\endgroup$ – TRUSKI Aug 2 '17 at 6:22
  • $\begingroup$ But the function can be from the set of irrational numbers also there it is not connected. If it is connected then it is not a problem. The only problem arises when it is disconnected. $\endgroup$ – Sachchidanand Prasad Aug 2 '17 at 6:23
  • $\begingroup$ Hmm... I don't the continuity is required for the first but I think it is for the second. Obviously if the continuous condition of the second were removed it wouldn't be true. (f(x) = 0 if x is rational, f(x) =1 if is irrational has a finite image). See what is used in the definition of continuity that makes thinks this a clinch. Maybe try for a a Cantor diagonal type argument. $\endgroup$ – fleablood Aug 2 '17 at 6:28
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There are non-constant continuous functions with an uncountable number of zeroes.

My first thought was to use the Cantor set $C$, and a search for prior art led to this existing example already on this site:

Non-constant continuous function having uncountably many zeros?

The function is:

$$f:[0,1]\to\mathbb{R}, f(x) = \inf_{c \in C}\{ |x - c|\}$$

It can be shown that $f(x) = 0 \quad\forall x \in C$ and that $f$ is continuous on $[0,1$].

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  • $\begingroup$ If $f(x)=0\forall x\in C$, then $f$ is a constant function on $C$, this it is not a counterexample for $2$. $\endgroup$ – 5xum Aug 2 '17 at 6:39
  • $\begingroup$ @5xum: Umm. $f(x) \ne 0$ for $x \notin C$. So $f$ is not constant on $[0, 1]$ and $f(C) = \{0\}$. $\endgroup$ – Aryabhata Aug 2 '17 at 6:40
  • $\begingroup$ OK, but then how is the function a counterexample? If you take $f$ defined on $\mathbb R$, then $f(\mathbb R)$ is uncountable - which makes it not a counterexample. $\endgroup$ – 5xum Aug 2 '17 at 6:42
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    $\begingroup$ @5xum: This negates the statement that "image on any uncountable set is uncountable under a non-constant continuous map". The shows an $f$ and an uncountable set $C$ for which this statement is not true. $\endgroup$ – Aryabhata Aug 2 '17 at 6:43
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For 1., you do not need continuity.

For $2$, the function $f: (-1, 0)\cup (0,1)$ defined as $$\begin{cases}-1 & x<0\\ 1 & x>0\end{cases}$$

is continuous on $(-1,0)\cup(0,1)$ and not constant.

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To 1, you dont need suppose continuous.

To 2, as the function is not constant, use intermediate value Theorem and get an interval in the image.

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    $\begingroup$ As I have mentioned earlier also that intermediate value theorem is for the connected set. It is not true for a disconnected set. And an uncountable set can be disconnected also, for example, $\mathbb{Q}^c$. $\endgroup$ – Sachchidanand Prasad Aug 2 '17 at 6:28

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