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So here is my question: I need to prove that a continuous function $f: M \mapsto \mathbb{Z}$, is constant provided that M is connected.

I am having trouble understanding this statement; if I set M = $\mathbb{R}$, how is $f$ constant?

Am I misinterpreting the question?

Thanks.

Community
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r123454321
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4 Answers4

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Hint: A singleton in $\mathbb{Z}$ is both open and closed. If you take a pre-image of a singleton in the range of $f$, what do you conclude?

Prahlad Vaidyanathan
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3

The image of a connected set under a continuous function is connected. What are the connected subsets of the integers?

lhf
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2

Claim:

a continuous function $f:\mathbb{R} \to \mathbb{Z}$ must be constant function.

If not, let ${\rm Image}(f)= \{r_{1},\ldots,r_{n},\ldots\}$ . Then $\exists $ disjoint open set $V_{1},\ldots,V_{n},\ldots$ s.t. $r_{1} \in V_{1},\ldots,r_{n}\in V_{n},$ and so on. Now $f^{-1}({V_{1}}),\ldots ,f^{-1}({V_{n}})$ all are disjoint open sets in $\mathbb{R}$. And their union is $\mathbb{R}$; which is a contradiction: as $\mathbb{R}$ is connected set. Hence ${\rm Image}(f)$ must be a singleton set.

Boris Novikov
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jon
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If f : M → R is continuous and M is connected, then f(M) is connected. Hence, assuming M is nonempty, f(M) must be a singleton or an interval.

Since neither the integers nor the irrationals contain an interval, if f(M) is a subset of either of them, f(M) must be a singleton, hence f must be constant.

skatkat915
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