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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 ?

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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 $(\mathbb{R},\sigma)$.

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

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  • $\begingroup$ Actually, $f^{-1}\bigl((-\varepsilon,\varepsilon)\bigr)=(-\infty,0)$. $\endgroup$ Apr 23, 2018 at 20:35
  • $\begingroup$ Right argument, wrong set. Sorry! $\endgroup$
    – giobrach
    Apr 23, 2018 at 20:37

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