Re: continuity of a function

Quote:

Originally Posted by

**student2011** In order to prove that a function is continuous, we need to check that the pre-image of any open set in B is open in A. I confused because there is no open set in B here, so what we can say, is it true that the function f is vacuously continuous.

The induced topologies on $A$ and $B$ are respectively $\mathcal{P}(A)$ and $\mathcal{P}(B)$ (parts of $A$ and $B$), so $f^{-1}(G)$ is open, for all $G$ open. This implies that $f$ is continuous.

Re: continuity of a function

Yes, I believe that's correct - all subsets of A and B are open since they are the intersection of A (or B) and an open set of real numbers. Therefore every function is continuous.

- Hollywood

Re: continuity of a function

Thank you so much. that means every subset of A or B is open because the intersection between any open set of real numbers with A or B is either empty or a singelton.

Thank you so much again

Re: continuity of a function

Not exactly. Every subset of A or B is open because every subset of A or B is the intersection of an open set of real numbers and A or B. So for example {2,4} is the intersection of A and the open interval (1.5,4.5) (or the open set $\displaystyle (1.99,2.01) \cup (3.99,4.01)$).

- Hollywood