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Consider an injective function $\,\,f:[0,1]\rightarrow[0,1].\,$ Then is it true that there always exists some non-empty open subinterval of $[0,1]$, such that $f([0,1])$ is dense in that subinterval? That is to say, do there exist $a$ and $b$ such that $a<b$ and for any $c$, $d$ in $(a,b)$, there exists a $y$ in $(c,d)$ such that $y=f(x)$ for some $x$?

I'd be much obliged if someone could give me ideas on how to go on about proving this or maybe provide a counter-example since I don't even know whether it's true or not.

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  • $\begingroup$ To be clear, $f$ is not required continuous? $\endgroup$
    – Seth
    Commented Oct 18, 2014 at 18:13
  • $\begingroup$ Nope, $f$ can be anything. All it has to be is injective. $\endgroup$
    – sayantankhan
    Commented Oct 18, 2014 at 18:15

1 Answer 1

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If $f$ is continuous, then the answer is: Yes.

If $f(x_1)=y_1<y_2=f(x_2)$, then $[y_1,y_2]\subset f[x_1,x_2]$, due to Intermediate Value Theorem.

However, if we allow $f$ to be discontinuous, then the answer is: No.

There exists an 1-1 and onto mapping between $[0,1]$ and the Cantor set, which is nowhere dense, and is equinumerous to $[0,1]$.

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  • $\begingroup$ My function $f$ need not be continuous. $\endgroup$
    – sayantankhan
    Commented Oct 18, 2014 at 18:16
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    $\begingroup$ See new version of my answer. $\endgroup$
    – Yiorgos S. Smyrlis
    Commented Oct 18, 2014 at 18:19

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