If $f:\mathbb R\to \mathbb N$ is continuous, then it's constant.

Question

I have to prove that if $f:\mathbb R\to \mathbb N$ is continuous, then it's constant. Th result looks intuitive, but I can't prove it. I try by contradiction : suppose there is $x,y\in\mathbb R$ s.t. $f(x)\neq f(y)$. So in somehow $|f(x)-f(y)|\geq 1$, and thus, using the $\varepsilon -\delta $ definition with $\varepsilon <1$ I tried to get a contradiction, but since there is no reason to have $|x-y|\leq \delta $, I couldn't find a way to write it properly. Can someone give hint ?

Do you know the Intermediate Value theorem ? – TheSilverDoe Nov 14 '20 at 19:49 — TheSilverDoe, Nov 14 '20 at 19:49
Also: https://math.stackexchange.com/q/505472/42969 – Martin R Nov 14 '20 at 19:52 — Martin R, Nov 14 '20 at 19:52

Surb · Answer 1 · 2020-11-14T19:50:15.853

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Method 1

Using your method : Suppose $x<y$ and $f(x)\neq f(y)$. Set $z=\inf\{t>x\mid f(t)\neq f(y)\}$. Using continuity of $f$ at $z$ should allow you to get a contradiction.

Method 2

$f^{-1}(\{f(0)\})$ is open and closed in $\mathbb R$.

edited Nov 14 '20 at 19:50

answered Nov 14 '20 at 19:48

Surb

55,662

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Method 3 : Use the IVT ! (+1) – TheSilverDoe Nov 14 '20 at 19:49
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Method 2 is the best one IMO, as that users the definition directly and it is still the shortest. When that happens it should be savoured. – Arthur Nov 14 '20 at 19:51
@Arthur: What mean IMO ? – Surb Nov 14 '20 at 19:53
@Surb "in my opinion" – Arthur Nov 14 '20 at 19:53
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I agree with you, but it's not sure that it's a good hint for the OP (depending on his/her topological knowledge) :-) @Arthur – Surb Nov 14 '20 at 19:55
I don't understand the 2nd method but thanks to method 1, I could solve my exercise. Thank you. – user841366 Nov 14 '20 at 20:35

If $f:\mathbb R\to \mathbb N$ is continuous, then it's constant.

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