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This is exercise $19$ from fourth chapter of baby Rudin.

Suppose $f$ is a real function with domain $R^1$ which has IVP i.e. if $f(a) < c < f(b)$, then $f(x)=c$ for some $x$ between $a$ and $b$. Suppose also for every rational $r$, that the set of all $x$ with $f(x)=r$ is closed. Prove that $f$ is continuous.

I use the hint given by author and got the answer. You can look here.

When I was thinking about this problem, I drew some graphs. One of the graph I draw has puzzled me. It seems to satisfy both the hypotheses, but fails to be continuous. The graph is as follows. enter image description here

Graph seems to satisfy IVP and also any horizontal line $y=k$ will hit graph at most twice, so second hypotheses also seems to satisfy. But $f$ is not continuous at origin. What I am messing up here. Thanks.

ogirkar
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Your example does not satisfy IVP for $[-x_{0},x_{0}]$ and $c=2$ if $x_{0}$ is small enough.

Salcio
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It does not satisfy the intermediate value property: $f(0)=3$ and $f(-1)<2$, but there is no $a\in(-1,0)$ such that $f(a)=2$.

José Carlos Santos
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  • I overlooked the condition $f(x)=c$ for some x between a and b. Thanks. Just a small question. Was there anything special about rationals in second hypothesis or we can just take any dense subset of $\mathbb{R}$ – ogirkar Jan 30 '21 at 11:43
  • Any dense subset of $\Bbb R$ will do. – José Carlos Santos Jan 30 '21 at 12:02