It is known that for a function $f$ from a topological space to interval $[0;1]$ to be continuous, it is enough that preimages $f^{-1}]a;1]$ and $f^{-1}[1;a[$ be open in $[0;1]$ for every $a$ in our interval.

Now let a function $f$ is from a pretopological space to interval $[0;1]$. What conditions are sufficient for $f$ to be continuous?

  • I might be wrong and i hope you have the patience to enlighten me,in your given definition,taking the space with usual topology $[-1,1]$ and $f(x)=x^{2}$ makes $f^{-1}([a,1[)=]-1,a]\cup[a,1[$ which is not an open set,even when $f$ is continuous. For the latter question it may be good to take the good old characterization of continuous as "pre-image of open set is an open set" – AHandsomeAlien Jul 11 '16 at 4:09
  • @AHandsomeAlien Certainly a typo. The question is corrected – porton Jul 11 '16 at 9:43
  • Now $[1,a[ = ]a,1]$ because $a\in[0,1]$ isnt it? If so,once again the function i gave you contradicts.because $f^{-1}(]a,1])=[-1,-a[\cup]a,1]$,no? – AHandsomeAlien Jul 11 '16 at 9:51
  • It is open in $[0;1]$ (despite of being not open in $\mathbb{R}$) – porton Jul 11 '16 at 12:12

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