Thinking of a function that is continuous only at one point

Question

Create an $f(x)$ that is continuous only at $x = 0$.

In other words, $f(x)$ is discontinuous at every other point. Maybe they are scattered points. However, for $f(x)$ to be continuous at $x = 0$, its limit as $x$ approaches zero must be $f(0)$. But even then, I cannot imagine such a function.

Here is another approach. Consider the function $f(x) = (cx)^n$. If $c$ and $n$ are very large, it acts like two vertical asymptotes squeezing from the sides very near $x = 0$, so that $f(x)$ is continuous only at $x = 0$.

Is my thinking correct? Are there any other ways that we can create such a function?

And perhaps this for a general observation. – Nov 23 '16 at 15:58 — , Nov 23 '16 at 15:58

score 2 · Accepted Answer · answered Nov 23 '16 at 15:52

2

Hints:

Note that if $f$ is bounded, then the function $x \mapsto x f(x)$ is continuous at $x=0$.

The function $f(x) = 1_\mathbb{Q}(x)$ is bounded and continuous nowhere.

answered Nov 23 '16 at 15:52

copper.hat

178,207

score 0 · Answer 2 · answered Nov 23 '16 at 16:09

0

$f(x)= \begin{cases} x &\text{$x\in \Bbb Q$}\\-x &\text{$x\in \Bbb Q^c$}\end{cases}$

answered Nov 23 '16 at 16:09

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Thinking of a function that is continuous only at one point

2 Answers2