# Determine if the following function is continuous?

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 ?

• Are you French? Apr 23, 2018 at 20:38

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 $(\mathbb{R},\sigma)$.
• i found in a book that $f^{-1}(]-\varepsilon,+\varepsilon[)=]-\infty,\sqrt{\varepsilon-1}[$ how they do? please Apr 23, 2018 at 20:50
• @Vrouvrou That's only when $\varepsilon>1$. Apr 23, 2018 at 20:52
• what about $\varepsilon<1$ please Apr 23, 2018 at 21:18
• @Vrouvrou If $x\geqslant0$, then $f(x)=x^2+1>\varepsilon$. Since, on the other hand,$$x<0\implies f(x)=0\in(-\varepsilon,\varepsilon),$$ $f^{-1}\bigl((-\varepsilon,\varepsilon)\bigr)=(-\infty,0)$. Apr 23, 2018 at 21:26
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 $(-\infty,0) \notin \sigma$.
• Actually, $f^{-1}\bigl((-\varepsilon,\varepsilon)\bigr)=(-\infty,0)$. Apr 23, 2018 at 20:35