Ytukyg

I'm looking for a counterexample to the statement:

If a function is continuous in a point $x_0$ then it is defined in a
neighborhood of that point.

If I take

$f: mathbb{R} setminus mathbb{Q} rightarrow mathbb{R}$ defined as $f(x):=x$

Then it is continuous in any point $x_0 in mathbb{R}setminus mathbb{Q}$ since $lim_{xto x_0} f(x)=x_0 forall x_0 in mathbb{R}$, nevertheless it is not defined in any neighbourhood of any point $x_0 in mathbb{R}$.

Is this counterexample correct?

Edit

Definition of neighborhood of $x_0$: Let $x_0 in mathbb{R}$ and $r>0$ a real number. We call a neighborhood of $x_0$ with radius $r$ the open and bounded interval in $mathbb{R}$ $(x_0-r,x_0+r)$.

Definition of limit: Let $x_0$ be a limit point for a function $f(x)$. We write $lim_{x to x_0} f(x)=l iff forall epsilon >0 exists delta >0 mid forall xin domf wedge 0<mid x-x_0 mid <delta implies mid f(x)-lmid <epsilon$

Definition of continuity of $f$ in $x_0$: $f$ is continuous in a point $x_0$ of its domain iff $forall epsilon >0 exists delta>0 mid forall x in dom(f) wedge mid x-x_0 mid < delta implies mid f(x)-f(x_0)mid <epsilon$

Assuming you are using the following definition your counterexample is correct. You could take $f(x)=0$ for all $xin mathbb{Q}$ for an even simpler example with a dense domain.

Let $x in X$. If $f:X to Y$ then we say that $f$ is continuous at $x$ if for all $varepsilon>0$ exists $delta>0$ such that

$$d_Y(f(x),f(y)) < varepsilon$$

whenever
$$yin X, ~ d_X(x,y) < delta$$

Note that this definition is equivalent to yours when $x$ is a limit point of the domain.

If $f$ is continuous at $x_0$ then $f^{-1}(mathrm{open})$ is open. So $f$ has to be defined in a neighbourhood of $x_0$.

No, this is not a valid counterexample because $f$ is not continuous anywhere which I will show as follows.

$$lim_{xto x_0}f(x)=f(x) iff forall varepsilon > 0, exists delta >0: [x in V_delta(x_0) implies f(x) in V_varepsilon(x_0)]$$

(where $V_a(b)$ is the open $a$-neighbourhood of $b$)

For any $varepsilon,delta>0$, I can obviously pick for $x$ any rational $q in V_delta(x_0)$, in which case $f(x)$ will not be defined and thus cannot be an element of $V_varepsilon(x_0)$.