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1.$f(x)$ is on $[0,1]$

2.for any $(\alpha,\beta)\subset [0,1]$, the discontinuous point set and the continuous point set of $f(x)$ are both uncountable.

i.e in any open set in $[0,1]$, there are uncountable continuous points and discontinuous points.

Any hint will be appreciated.

z3wood
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2 Answers2

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EDIT: I was extremely foolish earlier - my construction worked, but my verification was nonsense. Fixed!

Call a sequence of sets $A_i\subseteq [0, 1]$ ($i\in\mathbb{N}$) a good sequence if

  • Each $A_i$ is nowhere dense.

  • The $A_i$s are disjoint.

  • The set $B=\bigcup_{i\in\mathbb{N}} A_i$ is uncountably dense in $[0, 1]$: for any $0<\alpha<\beta<1$, $B\cap (\alpha,\beta)$ is uncountable.

Exercise: show that a good sequence exists.

Now given a good sequence, there is an associated function $f$, given by $f(x)=n$ if $x\in A_n$, and $f(x)=0$ if $x\not\in B$. I claim this function has the desired property.

First, we'll show that $f$ is continuous on $[0, 1]\setminus B$. We can show that $\lim_{x\rightarrow a}f(x)=0$ for all $a\in[0, 1]\setminus B$, since each $A_i$ is closed, as follows: fix $a\in [0, 1]\setminus B$ and $\epsilon>0$. Taking $n>{1\over \epsilon}$ and using the fact that finite unions of closed sets are closed, we may find an open interval $I\ni a$ which doesn't intersect $A_1\cup ... \cup A_n$; on this interval, $f$ is always $<\epsilon$. So $\lim_{x\rightarrow a}f(x)=0$.

Now since $f(a)=0$ for $a\not\in B$, this means that $f$ is continuous on $[0, 1]\setminus B$.

Next, we'll argue that $[0, 1]\setminus B$ is uncountably dense. This follows from the fact that the $A_i$s are each nowhere dense, using the Baire category theorem. Suppose otherwise; then for some $0\le\alpha<\beta\le1$, $B\cap [\alpha, \beta]$ must be countable. But

  • $B$ is meager,

  • any countable set is meager, and

  • the union of two meager sets is meager.

So $[\alpha, \beta]$ is meager; but this contradicts the Baire category theorem.

So the set of points of continuity of $f$ contains $[0, 1]\setminus B$, and hence is uncountably dense; it only remains to show that the set of points of discontinuity of $f$ is uncountably dense.

But this follows easily from the result above, and the easy fact that if a function $g(x)$ is constant (say, $=c$) on some dense set $D$ and $g(a)\not=c$ then $g$ is not continuous at $a$. So $f$ is discontinuous at every point in $B$. Since we already know by choice of $A_i$s that $B$ is uncountably dense, this means that the set of points of discontinuity of $f$ is uncountably dense (and with that we've used the final property of good sequences), so we are done.

Noah Schweber
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  • Thanks. But the set of discontinuities of $f$ is not infinite, right? – z3wood Jun 28 '17 at 14:48
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    In fact, most derivatives simultaneously have a $c$-dense continuity set and a $c$-dense discontinuity set. – Dave L. Renfro Jun 28 '17 at 14:49
  • @Goodboy Yes, it is. Every rational is a point of discontinuity of $f$ - this follows from the fact that $\lim_{x\rightarrow a}f(x)=0$ for every $a$. (HINT: for each $n$, there is some interval around $a$ not containing any rational with denominator $\le n$, hence not containing any number with $f(x)\ge{1\over n}$ ...) And of course there are infinitely many rationals in $[0, 1]$. – Noah Schweber Jun 28 '17 at 14:49
  • If the set of indexes $i$ of $A_i$ is not countable we need also uncountable non-zero values to replace the $1/n$. This forces them to accumulate also at points other than zero. This breaks the claims of continuity in the complement of the $A_i$. Yes, I did downvote. – OR. Jun 28 '17 at 14:55
  • Nowhere denseness has nothing to do with not accumulating. If you put an infinite set in a bounded interval it accumulates somewhere. If you put an uncountable set in a bounded interval it accumulates in many places. – OR. Jun 28 '17 at 15:07
  • I know all of that. Yes, I misread it as implying that the indexes could be made infinite. You want to put $A_i$ infinite. – OR. Jun 28 '17 at 15:14
  • @MariePierredeLeTetou I'm quite sorry - your criticism was absolutely right (this is what I get for doing math while travelling). The construction in my answer, however, works exactly as desired - if the $A_n$s are picked as described, the function $f$ is continuous exactly at the points not in $\bigcup_{n\in\mathbb{N}} A_n$, and both $\bigcup_{n\in\mathbb{N}} A_n$ and its complement are uncountably dense. I don't have time at the moment, but I'll fix this answer later tonight. – Noah Schweber Jul 01 '17 at 03:49
  • @MariePierredeLeTetou Fixed! – Noah Schweber Jul 01 '17 at 05:40
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Let $K$ be the Cantor set. By "a Cantor set" I'll mean any set of the form $a+bK,$ where $a\in \mathbb R, b>0.$

Let $I_1,I_2, \dots$ be the open subintervals of $[0,1]$ with rational end points. We can inductively choose pairwise disjoint Cantor sets $K_n,$ with $K_n\subset I_n$ for each $n.$

For each $n$ there is a countable set $D_n\subset K_n$ that is dense in $K_n.$ Define $f_n = 1/2^n$ on $D_n,$ and $f_n=0$ everywhere else on $[0,1].$ Then $f_n$ is continuous on each point of $[0,1]\setminus K_n.$ Furthermore the restriction of $f_n$ to $K_n$ is discontinuous at each point of $K_n\setminus D_n.$

We can now define

$$\tag 1 f=\sum_{n=1}^{\infty}f_n.$$

Note that for each $n,$ $f=f_n$ on $K_n.$

Because $0\le f_n\le 1/2^n,$ the series in $(1)$ converges uniformly on $[0,1].$ The continuity of $f_n$ on $[0,1]\setminus K_n$ shows that each $f_n$ is continuous on the set $U=[0,1]\setminus (\cup_{m=1}^{\infty}K_m).$ Therefore, by uniform convergence, $f$ is continuous on $U.$ Note that $\mu(U) = 1,$ since each $K_m$ has measure $0.$ (Here $\mu$ is Lebesgue measure.)

So now suppose $(a,b)$ is a subinterval of $[0,1].$ Then $(a,b)$ contains some $I_n.$ At each point of the set $(a,b)\cap U,$ $f$ is continuous. This set is uncountable, since the measure of this set is $b-a.$ Now $K_n\subset I_n,$ and since $f=f_n$ on $K_n,$ $f$ is discontinuous at each point of $K_n\setminus D_n$ as mentioned above. Because $K_n$ is uncountable and $D_n$ is countable, $K_n\setminus D_n$ is uncountable. This proves $f$ has the desired properties.

zhw.
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