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Let $f:[0,1]\to(0,1)$ be a surjective function. Prove that $f$ has at least one point of discontinuity.

$f$ is surjective, so all elements in codomain must be mapped to from some element in domain. I'm trying to think of some argument from here, but I'm stuck. Any hints please?

Ellie_Wong
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    The basic result is: the image of a compact space under a continuous function is compact. Since $[0,1]$ is compact, and $(0,1)$ is not, there can be no continuous surjection. – Thomas Andrews May 30 '23 at 20:33

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The interval $[0,1]$ is compact, so if $f$ had no discontinuities (i.e., if $f$ were continuous), then $f([0,1])$ would have to be compact. But by assumption, $f([0,1]) = (0,1)$, which is not compact, contradiction.

Nick F
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For each $n\in\Bbb N,$ choose $x_n\in[0,1]$ such that $f(x_n)=\frac1n.$ As a bounded sequence, $(x_n)$ admits a convergent subsequence $x_{n_k}\to x,$ and $x\in[0,1]$ (since $0\le x_{n_k}\le1\;(\forall k)$). This $x$ is a point of discontinuity of $f,$ otherwise we would have $$0=\lim_{k\to\infty}\frac1{n_k}=\lim_{k\to\infty}f(x_{n_k})=f(x)\in\operatorname{im}f=(0,1),$$contradiction.

Anne Bauval
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If $f$ where continuous the image of $f$ would be compact(closed and bounded) , since $[0, 1]$ is a compact interval, but the image of $f$ is $(0, 1) $, since $f$ is surjective. $(0, 1) $ is open so que have a contradiction.

Guillermo García Sáez
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  • There is an argument missing to explain why $(0,1)$ is not compact. Saying that it is open (in $\Bbb R$ I presume) is not sufficient. – Anne Bauval May 30 '23 at 21:04
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If you don't know about compactness, you need the lemma:

If $f:[a,b]\to\mathbb R$ is continuous then there exists an $x_0\in [0,1]$ such that $f(x_0)=\sup_{x\in [0,1]} f(x).$

It follows immediately from this lemma.

Thomas Andrews
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