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For $f : X \to Y$, the following S1 and S2 are equivalent.

S1. $f$ is continuous on $X$

S2. if $A \in Y$ is open in $Y$, then $f^{-1}(A)$ is open in $X$

Since open interval is only a special case of open set, I guess there must be an example of a function $f$ that satisfies S2 only in "interval" case, and $f$ is not continuous (because otherwise, S2 must reduce to "interval" statement).

But I'm having hard time coming up with such function. Any help will be appreciated.

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    $\begingroup$ An open set is a union of open intervals, and the inverse image of a union of the union of the inverse images, so... $\endgroup$ Oct 8, 2018 at 18:08
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    $\begingroup$ On the other hand, it's easy to come up with examples of functions that are continuous, but the inverse image of an open interval is not necessarily an open interval. e.g. $f(x) = \sin x$, $A = (-\frac{1}{2}, \frac{1}{2})$. $\endgroup$ Oct 8, 2018 at 22:34

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There is another equivalent condition

S3. For a fixed base $\mathcal{B}$ of the topology on $Y$, for all $B \in \mathcal{B}$ : $f^{-1}[B]$ is open in $X$.

S2. implies S3. because basic open sets are open in particular.

S3. implies S2. because if $O$ is open in $Y$, $O = \bigcup \mathcal{B}'$ for some subfamily $\mathcal{B}'$ of $\mathcal{B}$ by the definition of a base.

But then $f^{-1}[O] = \bigcup \{f^{-1}[B] : B \in \mathcal{B}'\}$ which by S3. is a union of open sets in $X$, hence open, as required.

As open intervals form a base for the topology on $\mathbb{R}$, if a function has the property that inverse images of intervals are open (or even open intervals in particular) then $f$ is continuous.

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