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I am stuck on the following problem that says:

Let $\,\displaystyle f \colon [0,1] \to [0,1]$ be continuous and $\,f(0)=0,f(1)=1.$ Then $f$ is necessarily

  1. injective ,but not surjective

  2. surjective,but not injective

  3. bijective

  4. surjective

I have to determine which of the following options is correct. Can someone help? Thanks and regards to all.

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  • $\begingroup$ You can start by drawing possible paths from $(0,0)$ to $(1,1)$ strictly contained in the unit square. $\endgroup$ – Pedro Tamaroff Oct 6 '13 at 17:42
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    $\begingroup$ Can you think of any such functions $f$? That'd be a good place to start... $\endgroup$ – Zev Chonoles Oct 6 '13 at 17:42
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you can kick out 1. and 2. by considering $f(x)=x$. furthermore A function with a graph like the letter $N$ will be not injective. But the function is surjective due to intermediate value theorem.

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  • $\begingroup$ @learner The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is a continuous function, then for every $\gamma \in [f(a),f(b)]$ there is $\xi \in [a,b]$ such that $f(\xi)=\gamma$. $\endgroup$ – Dominic Michaelis Oct 31 '13 at 20:34

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