If f,g cont. , then g o f cont.?

[Unassuming]

If f,g cont. , then "g o f" cont.?

Homework Statement

Let X,Y,Z be spaces and f:X-->Y and g:Y-->Z be functions.
If f,g are continuous,
then so is g o f.

Homework Equations

The Attempt at a Solution

This is my proof so far. I would be appreciative if someone could point out the weak points.

"Proof". Let f,g be continuous. Show "g o f" continuous. Show g(f(x)) continuous.

Since f is continuous, and since f:X-->Y, we know that x is an element of W such that f[W]$$\subseteq$$V for V open, V$$\subseteq$$Y.

Since g is continuous, and since g:Y-->Z, we know that f(x) is an element of V such that g[V]$$\subseteq$$M, for M open, M$$\subseteq$$Z.

Therefore, g[ f(x) ] $$\in$$ M $$\subseteq$$ Z.

Thus, "g o f" is continuous.

 

[cellotim]

Let me try to reiterate it. Say we have a neighborhood of f(x), V. Then we can find a neighborhood of x, U in X such that f(U) is in V. In the same way, if we have a neighborhood of g(f(x)) W in Z, we can find a neighborhood of f(x) T in Y, such that g(T) is in W. So, we should be able to find a neighborhood of x, S in X such that g(f(S)) is a subset of W.

 

[Unassuming]

It sounds a lot better your way. I understand it more now.