Continuity of functions in topology I have this topology $\sigma$ on $\mathbb{R}^2$ given by the emptyset and all the union of $\Omega_r=\{(x,y)\in\mathbb{R}^2, x^2+y^2=r^2\}, r\geq0$
The question is to study the continuity of the following two functions given by 
$ \Psi:(\mathbb{R},\lvert \cdot \rvert)\to (\mathbb{R}^2,\sigma)\\ \qquad x\mapsto (x,a)$ and $\Phi:(\mathbb{R}^2,\sigma)\to (\mathbb{R},\lvert \cdot \rvert)\\ \qquad(x,y)\mapsto x$
where $a$ is fixed in $\mathbb{R}$.
For the function $\Phi$ to study the continuity at each point of it's domain of definition, i think that it is enough to that the inverse image of any open set is open 
But if i take $]\alpha,\beta[$ that $\Phi^{-1}(]\alpha,\beta[)=]\alpha,\beta[\times\mathbb{R}$ it is not open in $(\mathbb{R}^2,\sigma)$
 so $\Phi$ is not continuous at any point from $\mathbb{R}$
is it correct ?  
 A: Your argument (well, you don't prove that the inverse image is not open) shows that $\Phi$ is not globally continuous, but it does not study the pointwise continuity of $\Phi$, as you asked.
It's easy to see that each point has one minimal neighbourhood: the circle around the origin going through it, so:
A function $f(\mathbb{R}^2, \sigma) \to X$ is continuous at $(x,y)$ iff for every open set $O$ of $X$ with $f(x) \in O$ we have $f[B_r] \subseteq O$ where $r = \sqrt{x^2 + y^2}$. 
As $B_0 = \{(0,0)\}$ this implies that all $f$  are continuous at $(0,0)$.
But as to $\Phi$, this criterion shows that $\Phi$ is only continuous there.
Also a function $f:X \to (\mathbb{R}^2, \sigma)$ is continuous iff $f^{-1}[B_r]$ is continuous for all $r \ge 0$.
Let $x$ be a point in $\Psi^{-1}[B_r]$. This means that $x^2 + a^2 = r^2$, which means that $r^2 \ge a^2$ and $x \in \{-\sqrt{r^2-a^2},\sqrt{r^2-a^2}\}$.
So $\Psi^{-1}[B_r]$ can be empty, but if it is not, then it is a finite subset of $\mathbb{R}$, so never open. In particular $\Psi^{-1}[B_{a^2}] = \{0\}$ for all $a$. So whatever $a$ is, $\Psi$ is not continuous.
