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I want to do a proof by contradiction. You guys let me know if I goofing up.

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1 Answer 1

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Suppose $f$ is non-constant. Since $f$ is continuous, it satisfies the Intermediate value theorem [in the most general sense, $f$ satisfies the theorem as long as $f$ is a mapping from any connected space $M$ to $\mathbb R$].

Pick any two irrational numbers $a<b$ in the image of $f$. Since $f$ satisfies the Intermediate value theorem, then $f$ attains all the intermediate values from $[a,b]$. We know that between any two irrationals, lies a rational--a contradiction. $f$ has a value that is not irrational.

This means the assumption of $f$ being non-constant is false. QED

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Your reasoning is correct. Just a linguistic quibble - the word "consecutive" would mean that there are no irrational numbers between $a$ and $b$, whereas I suspect you just meant to say that $a<b$. –  Brad Nov 5 '12 at 2:37
First, it makes no sense to talk about consecutive irrational numbers: between any two irrational numbers there is another irrational number, so they can’t be consecutive. Apart from that, the basic idea is sound, provided that $M$ is given to be a subset of $\Bbb R$; if not, you can’t use the intermediate value theorem. –  Brian M. Scott Nov 5 '12 at 2:38
@BrianM.Scott The intermediate value theorem applies for any connected topological space $M$: the image of $M$ under a continuous map $M\to\mathbb{R}$ is connected, and every connected subspace of $\mathbb{R}$ is an interval. –  Brad Nov 5 '12 at 2:41
I don't think it's uncommon to refer to the more general result as "the intermediate value theorem", e.g. in James Munkres's book Topology it is referred to as such. –  Brad Nov 5 '12 at 2:46
@Brad: I consider that a minor solecism on his part; it’s not the only one, though the book is certainly one of the best available. But then I also think it silly to bother, as he does, with the special case of continuous functions to linearly ordered spaces, since it’s a trivial consequence of preservation of connectedness by continuous maps. –  Brian M. Scott Nov 5 '12 at 2:52

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