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Daniele Tampieri
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Suppose f:RR$f:\mathbf{R}\to\mathbf{R}$ is a function. Let S={x∈R|f is continuous at x}$S=\{x\in \mathbf{R}|f\text{ is continuous at }x\}$. Does S$S$ have any nice properties?

Here are some observations about what S$S$ could be:

  • S$S$ can be any closed set. For a closed set S$S$, let g$g$ be a continuous function whose vanishing locus is S$S$ (for example, you could take g(x)$g(x)$ to be the distance of x$x$ from S$S$ if S$S$ is non-empty). Then define f(x)=g(x) if x∈Q and f(x)=0 otherwise. Then $$ f(x) = \begin{cases} g(x) &\text{if }x\in \mathbf{Q}\\ 0 &\text{otherwise}. \end{cases} $$ Then the continuous locus of f$f$ is exactly S$S$.
  • S$S$ can be an open interval. For an open interval S$S$, define f(x)=0 if x∈S or x∈Q and f(x)=1 otherwise. $$ f(x) = \begin{cases} 0 &\text{if }x\in S \wedge x\in\mathbf{Q}\\ 1 &\text{otherwise}. \end{cases} $$ Then the continuous locus of f$f$ is exactly S$S$.
  • S$S$ can be the complement of any countable set. Let T={t1,t2,t3,...}$T = \{t_1,t_2,t_3,\ldots\}=\{t_i\}_{i\in\mathbf{N}}$ be a countable set, and let ∑ai$\sum_i a_i$ be some absolutely convergent series all of whose terms isare non-zero (like ai=1/2i$a_i=\frac{1}{2^i}$). Define
      f(x) = ∑i such that ti < x a_i.
    $$ f(x) = \sum_{i\text{ s.t. }t_i < x} a_i. $$ Then the continuous locus of f$f$ is exactly the complement of T$T$.

Here are some questions I'd like to know the answers to:

  • Can S$S$ be any open set?
  • Can S$S$ be non-measurable? (if f(x)=0$$ f(x) = \begin{cases} 0 &\text{if }x\in S\\ 1 &\text{otherwise}. \end{cases} $$ $f(x)=0$ if x∈S$x\in S$ and f(x)=1$f(x)=1$ otherwise, what will the continuous locus be?)

Suppose f:RR is a function. Let S={x∈R|f is continuous at x}. Does S have any nice properties?

Here are some observations about what S could be:

  • S can be any closed set. For a closed set S, let g be a continuous function whose vanishing locus is S (for example, you could take g(x) to be the distance of x from S if S is non-empty). Then define f(x)=g(x) if x∈Q and f(x)=0 otherwise. Then the continuous locus of f is exactly S.
  • S can be an open interval. For an open interval S, define f(x)=0 if x∈S or x∈Q and f(x)=1 otherwise. Then the continuous locus of f is exactly S.
  • S can be the complement of any countable set. Let T={t1,t2,t3,...} be a countable set, and let ∑ai be some absolutely convergent series all of whose terms is non-zero (like ai=1/2i). Define
      f(x) = ∑i such that ti < x a_i.
    Then the continuous locus of f is exactly the complement of T.

Here are some questions I'd like to know the answers to:

  • Can S be any open set?
  • Can S be non-measurable? (if f(x)=0 if x∈S and f(x)=1 otherwise, what will the continuous locus be?)

Suppose $f:\mathbf{R}\to\mathbf{R}$ is a function. Let $S=\{x\in \mathbf{R}|f\text{ is continuous at }x\}$. Does $S$ have any nice properties?

Here are some observations about what $S$ could be:

  • $S$ can be any closed set. For a closed set $S$, let $g$ be a continuous function whose vanishing locus is $S$ (for example, you could take $g(x)$ to be the distance of $x$ from $S$ if $S$ is non-empty). Then define $$ f(x) = \begin{cases} g(x) &\text{if }x\in \mathbf{Q}\\ 0 &\text{otherwise}. \end{cases} $$ Then the continuous locus of $f$ is exactly $S$.
  • $S$ can be an open interval. For an open interval $S$, define $$ f(x) = \begin{cases} 0 &\text{if }x\in S \wedge x\in\mathbf{Q}\\ 1 &\text{otherwise}. \end{cases} $$ Then the continuous locus of $f$ is exactly $S$.
  • $S$ can be the complement of any countable set. Let $T = \{t_1,t_2,t_3,\ldots\}=\{t_i\}_{i\in\mathbf{N}}$ be a countable set, and let $\sum_i a_i$ be some absolutely convergent series all of whose terms are non-zero (like $a_i=\frac{1}{2^i}$). Define $$ f(x) = \sum_{i\text{ s.t. }t_i < x} a_i. $$ Then the continuous locus of $f$ is exactly the complement of $T$.

Here are some questions I'd like to know the answers to:

  • Can $S$ be any open set?
  • Can $S$ be non-measurable? (if $$ f(x) = \begin{cases} 0 &\text{if }x\in S\\ 1 &\text{otherwise}. \end{cases} $$ $f(x)=0$ if $x\in S$ and $f(x)=1$ otherwise, what will the continuous locus be?)
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Anton Geraschenko
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Does the "continuous locus" of a function have any nice properties?

Suppose f:RR is a function. Let S={x∈R|f is continuous at x}. Does S have any nice properties?

Here are some observations about what S could be:

  • S can be any closed set. For a closed set S, let g be a continuous function whose vanishing locus is S (for example, you could take g(x) to be the distance of x from S if S is non-empty). Then define f(x)=g(x) if x∈Q and f(x)=0 otherwise. Then the continuous locus of f is exactly S.
  • S can be an open interval. For an open interval S, define f(x)=0 if x∈S or x∈Q and f(x)=1 otherwise. Then the continuous locus of f is exactly S.
  • S can be the complement of any countable set. Let T={t1,t2,t3,...} be a countable set, and let ∑ai be some absolutely convergent series all of whose terms is non-zero (like ai=1/2i). Define
    f(x) = ∑i such that ti < x a_i.
    Then the continuous locus of f is exactly the complement of T.

Here are some questions I'd like to know the answers to:

  • Can S be any open set?
  • Can S be non-measurable? (if f(x)=0 if x∈S and f(x)=1 otherwise, what will the continuous locus be?)