- #1
Robin04
- 260
- 16
Definition: A function f mapping from the topological space X to the topological space Y is continuous if the inverse image of every open set in Y is an open set in X.
The book I'm reading (Charles Nash: Topology and Geometry for Physicists) emphasizes that inversing this definition would not be a sufficient requirement for continuity: a function that maps open sets into open sets is not necesserily continuous (according to our intuition). He presents a counter example ## f(x) =
\begin{cases}
x& x\leq 0 \\
1+x & x > 0
\end{cases}
##
He says that this is a function that maps open sets into open sets, but the inverse image of an open set is not necesserily an open set. The second part is trivial if we look at the image of the interval ##(1- \epsilon, 1+\epsilon)##, so according to the real definition this isn't a continuous function. But I'm having a problem with the first part. If we take the interval ##(-\epsilon,\epsilon)##, then its image is ##(-\epsilon, 0] \cup(1,1+\epsilon)## This is not an open interval either. What am I missing here?
The book I'm reading (Charles Nash: Topology and Geometry for Physicists) emphasizes that inversing this definition would not be a sufficient requirement for continuity: a function that maps open sets into open sets is not necesserily continuous (according to our intuition). He presents a counter example ## f(x) =
\begin{cases}
x& x\leq 0 \\
1+x & x > 0
\end{cases}
##
He says that this is a function that maps open sets into open sets, but the inverse image of an open set is not necesserily an open set. The second part is trivial if we look at the image of the interval ##(1- \epsilon, 1+\epsilon)##, so according to the real definition this isn't a continuous function. But I'm having a problem with the first part. If we take the interval ##(-\epsilon,\epsilon)##, then its image is ##(-\epsilon, 0] \cup(1,1+\epsilon)## This is not an open interval either. What am I missing here?