# On continuous functions

• Apr 29th 2008, 09:20 PM
squarerootof2
On continuous functions
so i'm given that (S_1,d_1), (S_2,d_2), (S_3, d_3) are metric spaces and f_1: (S_1,d_1)->(S_2,d_2) and f_2: (S_2,d_2)->(S_3,d_3) are continuous functions. i need to prove that f_2 composed with f_1 is continuous.

so the definition i'm working is that this function will be continuous if for every open set in S_3, the set's inverse image is also open in S_1. would the right way to do it here be to show that (f_2 composed with f_1)^(-1)(U) = f_1^(-1)(f_2^(-1)(U)) first then use the fact that f_1 and f_2 are continuous functions? i'm not sure if what i'm trying to do is right. thanks for help.
• May 1st 2008, 12:58 AM
Opalg
Quote:

Originally Posted by squarerootof2
so i'm given that (S_1,d_1), (S_2,d_2), (S_3, d_3) are metric spaces and f_1: (S_1,d_1)->(S_2,d_2) and f_2: (S_2,d_2)->(S_3,d_3) are continuous functions. i need to prove that f_2 composed with f_1 is continuous.

so the definition i'm working is that this function will be continuous if for every open set in S_3, the set's inverse image is also open in S_1. would the right way to do it here be to show that (f_2 composed with f_1)^(-1)(U) = f_1^(-1)(f_2^(-1)(U)) first then use the fact that f_1 and f_2 are continuous functions? i'm not sure if what i'm trying to do is right. thanks for help.

Yes, that's exactly the right approach. If U is open in S_3, then $f_2^{-1}(U)$ is open in S_2 (because f_2 is continuous). Therefore $(f_2\circ f_1)^{-1})(U) = f_1^{-1}\bigl(f_2^{-1}(U)\bigr)$ is open in S_1 (because f_1 is continuous).