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I wish to know if the following is true:

Let $f : [\alpha, \beta]\to \mathbb R$ be a function so that $$ \lim_{x\to x_0} f(x) = 0$$ for all $x_0 \in [\alpha, \beta]$. Then $f(x) = 0$ for some $x\in [\alpha, \beta]$.

The Thomae's function $f: [0,1]\to \mathbb R$

$$f(x) =\begin{cases} 1/q & \text{if }x= p/q\in \mathbb Q, \\ 0 & \text{otherwise.}\end{cases}$$

leads me to the above question. The Thomae function has limit zero everywhere, althought it is nonzero in $\mathbb Q$. I think I can take any countable dense subset $D\subset [\alpha, \beta]$ and construct a function which is nonzero in $D$ but limit equals zero everywhere. But I can't think of a function that is nonzero everywhere but has zero limit everywhere.

Arctic Char
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1 Answers1

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The number of points $x\in[\alpha,\beta]$ at which $|f(x)|>1/n$, for a fixed $n\in\mathbb{N}$, has to be finite. Otherwise, since the interval is compact, they accumulate somewhere and at that point the limit wouldn't be zero.

Therefore, the points $x\in[\alpha,\beta]$ at which $|f(x)|\neq0$ is countable, but $[\alpha,\beta]$ isn't (unless $\alpha=\beta$ but that case follows directly).

user647486
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