# Does there exist a function such that $\lim_{x \to a} f(x) = L$ for all $a \in \mathbb R$ but $f(x)$ is never $L$?

Does there exist a function $f:\mathbb R \to \mathbb R$ such that $\lim_{x \to a} f(x) = L$ for all $a \in \mathbb R$ but $f(x) \neq L$ for all $x$?

I found such a function in $\mathbb Q \to \mathbb Q$, where $f(\frac{p}{q})= \text{first p+q digits of }\pi$ satisfies the above condition. However, I have not been able to extend it to reals. So, is such a function possible, and if it is, is there any explicit example preferably related to the above function in rationals?

• Another example on the rationals is $f(p/q)=L-(1/q)$. May 27, 2016 at 9:28
• It is instructive to consider what happens when the limit $L$ is not same for every point $a$. The conclusion in still true. See math.stackexchange.com/q/980022/72031 for more details. May 27, 2016 at 10:13
• The limit for points where your function is defined is the function value at that point, so how could it be unequal to all function values??? May 27, 2016 at 10:55
• @hkBst: The limit for points ... is not necessarily the function value at that point. Remember a limit is defined in terms of values of a function, but it may not itself be a value of the function. May 27, 2016 at 11:26
• also see this beautiful answer math.stackexchange.com/a/3802/72031 for the general problem mentioned in my previous comment. This answer avoids the notion of uncountability and instead relies of completeness of real number system. May 30, 2016 at 19:34

Let's consider an arbitrary closed interval $[a, b]$. Let $\epsilon > 0$ be arbitrarily given. For each point $c$ of $[a, b]$ there is a neighborhood $I_{c}$ of $c$ such that $$|f(x) - L| < \epsilon$$ for all $x \in I_{c} \setminus \{c\}$. Clearly all such intervals $I_{c}$ form an open cover for $[a, b]$ and by Heine Borel Theorem a finite number of such intervals say $I_{c_{1}}, I_{c_{2}}, \ldots, I_{c_{m}}$ cover $[a, b]$.

It thus follows that $|f(x) - L| < \epsilon$ for all $x \in [a, b]$ except for a finite number of points $c_{1}, c_{2}, \ldots, c_{m}$. Now we choose specific values of $\epsilon$. For each positive integer $n$ we have a finite number, say $k_{n}$, of points in $[a, b]$ for which $|f(x) - L| \geq 1/n$. Let the set of points in $[a, b]$ for which $|f(x) - L| \geq 1/n$ be denoted by $A_{n}$. Then $A_{n}$ is a finite set of cardinality $k_{n}$ and since the set of points in $[a, b]$ for which $f(x) \neq L$ is obviously contained in the union $\bigcup_{n = 1}^{\infty}A_{n}$ it follows that the set of points in $[a, b]$ for which $f(x) \neq L$ is countable. It follows that $f(x) = L$ on $[a, b]$ for uncountably many points $x$.

• Yeah your argument is much more creative (+1) May 27, 2016 at 10:08
• @Timkinsella: Thanks man! but I will do have a look on wiki for Baire Category just in order to understand your proof better. May 27, 2016 at 10:09

Suppose $f$ is such a function. Like @Tim kinsella did, set $A_n = \{x\in [0,1]: |f(x) - L| > 1/n\}.$ Then $[0,1] = \cup A_n.$ Hence some $A_{n_0}$ must be infinite. By Bolzano-Weierstrass, $A_{n_0}$ has an accumulation point $x_0.$ Thus $x_0$ is a limit of a sequence $x_m$ such that $|f(x_m) - L| > 1/n_0$ for every $m.$ Since $\lim_{x\to x_0} f(x) = L,$ we have a contradiction.

No. Suppose this is true of $f(x)$. Consider $A_n:= \{x: |f(x)-L|>1/n\}$. $\cup_n A_n=\mathbb{R}$ by assumption. If none of the $A_n$ had an accumulation point, then $\mathbb{R}$ would be a countable union of sets whose compliments are open and dense. Such a union must have open dense compliment by the Baire category theorem. This is a contradiction. Maybe theres a more elementary way to do it though.

As user254something points out, you don't need Baire category. Just observe that discrete implies countable. Then $\mathbb{R}$ is a countable Union of countable sets. Contradiction.

• I have tried to use similar argument in my answer via Heine Borel Theorem, but I am not familiar with Baire Category Theorem so can't really comment on your answer. May 27, 2016 at 9:59
• If $A_n$ is discrete then $A_n$ is countable, because $A_n \cap [m, m+1]$ must be finite for every $m\in Z.$ May 27, 2016 at 16:25
• @user254665 That's a good point. May 27, 2016 at 21:52
• @Tim kinsella, Can there exist $f:R\to R$ such that $L(x)=\lim_{y\to x, y\ne x}f(y)$ exists for all $x,$ but$f(x) \ne L(x)$ for all $x$? May 28, 2016 at 2:02
• @user254665 If there were such an $f$, then $f(x)-L(x)$ would be a function which is never 0 but has limit 0 at every point. May 28, 2016 at 2:37