A fellow member of the community asked: "there isn't a sequence of continuous function on $[0,1]$ that converges pointwise to the function $f$ on $[0,1]$ defined by $f(x)=0$ if $x$ is rational and $f(x)=1$ if $x$ is irrational." This is the reverse of $\chi_{\mathbb{Q}}$, but the proof is obviously the same.

I'm wondering about a solution that only uses the definition of continuity and point-wise convergence. I answered the OP with the following attempt:

Attempt 1: " Suppose that $\{f_{n}\}_{n=1}^{\infty}$ is a point-wise convergent sequence that converges to $f$. Let $x \in [0,1]$ and WLOG, suppose that $x$ is irrational.

Let $\epsilon>0$. WLOG, suppose that $\epsilon<1$.

1) Then there exists $N_{1} \in \mathbb{N}$ such that $b \in \mathbb{N}$ and $b>N_{1}$ imply that $|f_{b}(x)-f(x)|=|f_{b}(x)-1|<\frac{\epsilon}{2}$.

2) Since $\{f_{n}\}_{n=1}^{\infty}$ is a sequence of continuous functions, there exists some $\delta>0$ such that $|x-y|<\delta$ implies that $|f_{b}(y)-f_{b}(x)|<\epsilon$ for each $y \in [0,1]$. Let $y \in [0,1]$ such that $|x-y|<\delta$. Then choose $p$ such that $p \in \mathbb{Q}$, where $x<p<y$.

4) Since $\{f_{n}\}_{n=1}^{\infty}$ is pointwise convergent, there exists $N_{2} \in \mathbb{N}$ such that $a \in \mathbb{N}$ and $a>N_{2}$ imply that $|f_{a}(p)-f(p)|=|f_{a}(p)|<\frac{\epsilon}{2}$.

5) Let $N=\max\{N_{1},N_{2}\}$. Suppose that $n>N$. By hypothesis, $|f_{n}(p)|<\frac{\epsilon}{2}$ and $|f_{n}(x)-1|<\frac{\epsilon}{2}$. However, by (3), we know that $|f_{n}(x)-f_{n}(p)|<\epsilon$. But this is clearly a contradiction."

However, this doesn't work, since it assumes that the same $\delta$ will characterize continuity for any $f_{n}$ where $n>N$.

**as a side-remark, is there any literature on the idea of a single $\delta$ working to describe continuity in a sequence of point-wise convergent functions $f_{n}$ for sufficiently large $n$?

Attempt 3: sigh, here is another proof that ultimately rests on baire category for contradiction:

Let $\{f_n(x)\}$ be a sequence of continuous functions that converge pointwise.

We show that $A=\{x \in \mathbb{R} \mid f_n(x) \textrm{converges}\}$ is a Borel set [and more specifically, $G_\delta$.

If $x \in A$, then there exists $N \in \mathbb{N}$ so that $n,m \geq N$ implies that $V_{x,n,j}=|f_n(x)-f_m(x)|<\frac{1}{j}$ for each $j \in \mathbb{N}$. Then Let $V_{n,j}=\bigcup V_{x,n,j}$. Since $V_{n,j}$ is open, let the open set $U_{n,j}$ be defined by $U_{N,J}=F^{-1}(V_{n,j})$. But then $$\bigcap_{j \in \mathbb{N}} \bigcup_{N \in \mathbb{N}} \bigcap_{n,m \geq N} U_{n,j}=A$$ is $G_\delta$

Then use Baire category to show that $\mathbb{Q}$ is not $G_\delta$, and so the result follows.

I'm still interested in a proof that does not use this, and also in a possible proof verification for my proof given in the answers section.

  • 3
    $\begingroup$ Equicontinuity captures the notion of "the same $\delta$ working for every function." See here: en.wikipedia.org/wiki/Equicontinuity $\endgroup$
    – kccu
    Dec 30, 2015 at 5:22
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    $\begingroup$ I don't think the Baire Category Theorem is "unnecessary machinery" here. If $f$ is a pointwise limit of continuous functions then $f^{-1}\{0\}=\{x : f(x)=0\}$ is a $G_{\delta}$ set. The simplest way to show that $Q$ is not $G_{\delta}$ is to use Baire. $\endgroup$ Dec 30, 2015 at 7:03
  • $\begingroup$ I know it is the easiest and simplest way. However, I'm just wondering about an alternative method. $\endgroup$ Dec 30, 2015 at 7:06
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    $\begingroup$ Would this answer satisfy you? Other copies of this question might be worth checking, too. Some of them can be found here or here. $\endgroup$ Dec 30, 2015 at 9:11
  • $\begingroup$ @user254665 only now do I understand your concern for what is probably as simple a proof as one can muster up. For some reason, there is some part of me that still cannot let go of the fact that there is something more fundamentally wrong with the notion of this particular convergence. $\endgroup$ Mar 16, 2016 at 3:39

2 Answers 2


A way to show that $ Q$ is not $G_{\delta}$ in $R$ without the Baire Category theorem:

(1).A non-empty completely metrizable space with no isolated points is an uncountable space.(In fact it has a subspace homeomorphic to the Cantor set.)

(2).A subspace $Y$ of a completely metrizable space $X$ is completely metrizable iff $Y$ is $G_{\delta}$ in $X.$

Therefore, since the space $Q$ has no isolated points, it cannot be completely metrizable, else by (1) it would be uncountable. So by (2), $Q$ is not $G_{\delta}$ in $R.$

Both (1) and (2) are provable by elementary, but not brief, methods, without any reference to the Baire Category theorem.

  • $\begingroup$ This is a fascinating point! (1) is a theorem in Munkres' Topology. I will work on the second bit! Thank you $\endgroup$ Mar 17, 2016 at 3:27
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    $\begingroup$ The direction of (2) needed here, that if Y is completely metrizable then Y is G-delta in X, is easier than the other direction. There is another result proved very much like the proof of (1): If X is locally compact,Hausdorff, non-empty and has no isolated points then X has a compact subspace of cardinal at least |R| (the cardinal of the reals). $\endgroup$ Mar 17, 2016 at 9:03
  • $\begingroup$ I will accept this answer, as it does indeed answer the question, and it does not seem as though any new suggestions will come. $\endgroup$ Oct 27, 2016 at 8:21
  • $\begingroup$ (2) is best done in stages: 1. If $X$ is metrizable , $Y$ completely metrizable with $Y\subset X,$ then $ X$ \ $\bar Y$ is $F_{\sigma} in X.$ .And show that $\partial Y$ is $F_{\sigma}$ in $X$.... (2). If $X$ is Hausdorff and $F=\{f_n:n\in N\}$ is a set of completely metrizable subspaces of $X,$ let $d_n$ be a complete metric on $f_n$ with $\sup f_n\leq 1.$ Show that $\sum_{n\in N}2^{-n}d_n$ generates the subspace topology on $\cap F$ and is a complete metric. $\endgroup$ Oct 28, 2016 at 18:59

I think I have an answer, I received some assistance from a friend of mine. Slight question: does it depend implictly on any non-stated theorems (I mean, so called "big theorems.")

Suppose that $\{f_{n}\}_{n=1}^{\infty}$ is a point-wise convergent sequence that converges to $f$. Let $a,b \in [0,1]$ such that $0<a<b<1$. We will show that for each non-degenerate segment $A \in [0,1]$, there exists some $B \subseteq A$ and arbitrarily large $N \in \mathbb{N}$ such that $f_{N}(B) = [a,b]$.

Let $A$ be a non-degenerate segment of $[0,1]$. Let $y \in (\mathbb{R}-\mathbb{Q})\cap A$ and $x \in \mathbb{Q} \cap A$ such that $x < y$. Since $\{f_{n}\}_{n=1}^{\infty}$ converges pointwise to $f$, we have that there exists $N \in \mathbb{N}$ such that $f_{N}(x) \geq b$ and $f_{N}(y) \leq a$. But since $f_{N}$ is continuous,by the Intermediate Value Theorem, there exists some $ B \subseteq [x,y] \subseteq A \subseteq [0,1]$ such that $f_{N}(B)=[a,b]$.

Then consider the embedded sequence of $\{A_{n}\}$, where $A_{n+1} \subseteq A_{n}$. Clearly for each $A_{n}$ there exist respective $K_{n} \in \mathbb{N}$ such that $f_{K_{n}}(B_{n})=[a,b]$ for $B_{n} \subseteq A_{n}$. By the nested interval theorem, the intersection of $A_{n}$ is non-empty. Then let $x \in \bigcap_{n \in \mathbb{N}}A_{n}$. But then $f(x)$ is the limit of $f_{n}(x) \in [a,b]$, which is a contradiction since $f(x)=0$ or $f(x)=1$ and $0<a<b<1$.

**edit: i think this proof is incorrect since it makes little reference to the rationals (Although I suppose it assumes denseness.)

  • $\begingroup$ It IS correct, Consider that both $\mathbb Q$ and its complement are both dense. You can replace $\mathbb Q$ with any $S$ which is dense and co-dense. $\endgroup$ Oct 28, 2016 at 19:11
  • $\begingroup$ Well that is quite heartening! It has been a while since I thought about this, but retrospectively, I don't see anything wrong either. Interesting remark of yours, I suppose that dense and codense would he sufficient to generalize this proof. $\endgroup$ Oct 28, 2016 at 19:15

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