Determine if the following function is continuous?

Question

I want to study the continuity $$f(x)= \begin{cases} 0,~\text{if}~ x<0\\ x^2+1,~\text{if}~ x\geq0 \end{cases} $$ where $f:(\mathbb{R},\sigma)\to (\mathbb{R},|.|)$ $$\sigma=\{\emptyset\}\cup\{\Omega\subset\mathbb{R},~ \rm card(\mathbb{R}\setminus \Omega)<+\infty\}$$ is the co-finite topology,

i say let $x<0$ then $f(x)=0$, let $W=]-\varepsilon,+\varepsilon[$ how to find $f^{-1}(]-\varepsilon,+\varepsilon[)$ please ?

Answer 1

Let $0 < \varepsilon < 1$ and consider the set $(-\varepsilon, \varepsilon) \in |\cdot|$. The preimage of this set under $f$ is, by definition, the set of all points in the domain of $f$ that get mapped to $(-\varepsilon,\varepsilon)$. In this case, the all points in $(-\infty,0)$ get sent to $(-\varepsilon,\varepsilon)$, so $f^{-1}((-\varepsilon,\varepsilon)) = (-\infty,0)$. As you can see, the cardinality of $\mathbb R \setminus (-\infty,0)$ is not finite, so $(-\infty,0) \notin \sigma$.

Answer 2

Your function is not continuous, because $(-1,1)$ is an open subset of $\mathbb R$ (with respect to the usual topology), but $f^{-1}\bigl((-1,1)\bigr)=(-\infty,0)$, which is not open in $(\mathbb{R},\sigma)$.