Does such function exist?
- Grigorios Kostakos
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Does such function exist?
Does exist a real function \(f:[a,b]\longrightarrow{\mathbb{R}}\) which is bounded, monotonic and discontinuous in uncountable many points of \([a,b]\) ?
Grigorios Kostakos
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Re: Does such function exist?
No there isn't even if $f$ is allowed to be unbounded. So let $f$ be a monotonic function. We will show that it has at most countably many discontinuities. Let us also suppose without loss of generality that $f$ is monotone increasing.
Since $f$ is monotonic, then it has left and right limits at each $x \in (a,b)$. Let $x_{\ell},x_r$ be the corresponding limits. If $f$ is discontinuous at $x$, then $x_{\ell} < x_r$. Furthermore, since $f$ is increasing, then it cannot take any value in the interval $(x_{\ell},x_r)$. Let $q_x \in (x_{\ell},x_r) \cap \mathbb{Q}$.
So we have exhibited for each point of discontinuity $x$ of $f$ (except perhaps at $x=a$ and $x=b$) a rational number $q_x$. Furthermore we have $x < y \iff q_x < q_y$.
Thus $f$ can only have countably many discontinuities.
Since $f$ is monotonic, then it has left and right limits at each $x \in (a,b)$. Let $x_{\ell},x_r$ be the corresponding limits. If $f$ is discontinuous at $x$, then $x_{\ell} < x_r$. Furthermore, since $f$ is increasing, then it cannot take any value in the interval $(x_{\ell},x_r)$. Let $q_x \in (x_{\ell},x_r) \cap \mathbb{Q}$.
So we have exhibited for each point of discontinuity $x$ of $f$ (except perhaps at $x=a$ and $x=b$) a rational number $q_x$. Furthermore we have $x < y \iff q_x < q_y$.
Thus $f$ can only have countably many discontinuities.
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