So I adapted this argument from one that assumes a closed interval, but I can't see why the arguement fails (even though I think it should) if we're talking about an open interval

Theorem if $f$ is continuous on $(a,b)$, then $\exists y \in (a,b)$ such that $f(y) \geq f$

Let $A = \big \{ f(x) : x \in (a, b) \big \}$

$A$ is bounded, and obviously not empty. Hence there exists $\alpha = \text{sup} \ A$

Suppose $\alpha \not \in A$, and therefore $\alpha > f(x)$

Consider $(g \circ f) (x) = \dfrac{1}{\alpha - f(x)}$

We know $\alpha - f(x) \in \mathbb{P}$, therefore $g$ is cts. everywhere

Consider $\alpha - \epsilon < b$. It is certainly not an upper bound, and so there exists $f(x') \in A$ where $f(x') > \alpha - \epsilon$.

\begin{align*} &\phantom{\Rightarrow}\alpha - \epsilon < f(x') \\ &\Rightarrow \alpha - f(x') < \epsilon \\ &\Rightarrow \frac{1}{\epsilon} < \frac{1}{\alpha - f(x')} \\ &\Rightarrow \frac{1}{\epsilon} < (x') && \left (f(x') \in A \right) \\ &\Rightarrow N < g(x') \end{align*}

A contradiction, for $g$ is continuous on $(a,b)$ and therefore bounded above.

  • $\begingroup$ What does "cts.@" mean? $\endgroup$ – Saucy O'Path May 14 at 14:40
  • $\begingroup$ continuous - I'll make the correction $\endgroup$ – user_hello1 May 14 at 14:40
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    $\begingroup$ The function $f(x)=x$ is a counterexample on the interval $(0,1)$. $\endgroup$ – lulu May 14 at 14:42
  • $\begingroup$ @lulu lol fantastic counterexample. $\endgroup$ – Randall May 14 at 14:42
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    $\begingroup$ As I said, the function $f(x)=x$ is a counterexample to your theorem on $(0,1)$. That is, there is no $y\in (0,1)$ such that $f(y)≥f(x)$ for every $x\in (0,1)$. If you meant something else, please edit accordingly. Otherwise, I suggest going through your argument using my function to see where it goes wrong. $\endgroup$ – lulu May 14 at 14:50

Your set $A$ need not be bounded. For example, take $f(x) = \frac{1}{x}$ with domain $(0,1)$. In particular, the resulting set $A$ in this example has no upper bound, so you may not use its supremum as in your argument.


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