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Suppose that $f:\mathbb{R} \rightarrow \mathbb{R}$ is a continuous function. Prove that there exist intervals $I_1$ and $I_2$ such that

$\sup_{x \in I_{1}}f(x) \leq \inf_{x \in I_{2}}f(x)$

Now, we can consider any two intervals [a,b] and [c,d] so that $b \leq c$. Since f is continuous, we have that $\sup f(x)$ exists on [a,b] and similarly $\inf f(x)$ on [c,d]. If we get the above inequality then we are done. Otherwise, I tried to use the properties of sup and inf to get the inequality but the direction flips. I do not how to proceed further. Any hint would be appreciated.

Tim
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    I think you need to add more details to your question. Is f a continuous function from $\mathbb{R}$ to $\mathbb{R}$? If so, take an arbitrary interval $I=[a,b] \subset \mathbb{R}$ with $a<b$. If $f$ is a constant on $[a,b]$ we can select $I_1$ and $I_2$ as arbitrary disjoint subintervals of $I$. Otherwise $f$ attains its supremum and infimum on $I$, let us say at $x_s$ and $x_i$ with $x_s\neq x_i$. Take $I_1$ and $I_2$ to be sufficiently small disjoint intervals around $x_i$ and $x_s$, respectively. – rafexiap Jul 30 '23 at 07:26
  • Thanks, just edited. @rafexiap – Tim Jul 30 '23 at 07:29

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Hint: If $f$ is a constant on $[0,1]$ there is nothing to prove. Otherwise, there is point $c$ where $f$ attains its minimum on $[0,1]$ and a point $d$ where $f$ attains its maximum on $[0,1]$. Note that $f(c)<f(d)$. Consider a small interval $I_1$ around $c$ and a small interval $I_2$ around $d$.

geetha290krm
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