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How would I prove that if I have a function $f$ on $\mathbb R^1$, that is continuous and non constant, that its range is non countable?

Here's my thought.

Let $f$ be a continuous, non constant function on $\mathbb R^1$. Because $f$ is non constant, it is not connected. Therefore, there exists a subset of $\mathbb R^1$ that is both open and closed within the preimage of $f$. Therefore, either the domain or range of $f$ is disconnected. Thus there is not always an intermediate value on all intervals in $\mathbb R^1$...

Am I close at all?

Joseph Paul
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  • use the intermediate value theorem –  Oct 11 '14 at 01:48
  • Can I do this as a consequence of the metric space not being connected? – Joseph Paul Oct 11 '14 at 01:51
  • Sure, can you prove that a countable subset of $\mathbb R$ having at least two points is disconnected? – Erick Wong Oct 11 '14 at 01:56
  • I thought it was disconnected by virtue of being nonconstant. Thus, either the domain or range is also disconnected. Therefore, it wouldn't take on all values of f (a) and f (b). Hence there would be no median value? – Joseph Paul Oct 11 '14 at 02:01
  • Nonconstant just means there exist $a, b \in \mathbb R$ so that $f(a) \neq f(b)$. A constant function is one of the form $f(x) = c$ for all $x \in \mathbb R$, with $c \in \mathbb R$ fixed. It's not clear what you mean by $f$ being disconnected, but it certainly isn't standard terminology. – Dustan Levenstein Oct 11 '14 at 03:01
  • That makes sense! Thanks! – Joseph Paul Oct 11 '14 at 03:03
  • F being connected was a phone typo. I meant the metric. But how would I go on to prove the range is uncountable? – Joseph Paul Oct 11 '14 at 03:04
  • @JosephPaul glad to hear it! I'll post an answer addressing the actual question. – Dustan Levenstein Oct 11 '14 at 03:08

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Hint: let $a<b$ be two points so that $f(a) \neq f(b)$. Then either $f(a)<f(b)$ or $f(a)>f(b)$. Let's discuss the case $f(a)<f(b)$, and the other case can be handled similarly.

Let $c \in \mathbb R$ be any number in the range $f(a)<c<f(b)$. Note that there are uncountably many such $c$. Now what does the intermediate value theorem tell you?

Dustan Levenstein
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$f$ is non-constant. hence range of f contains more than one element. R is connected and $f $ is continuous.hence $f(R)$ is connected.It contains at least two distinct points hence $f(R)$ is uncountable as theorem says that connected set with two distinct points is uncountable.

ASHWINI SANKHE
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