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

  • Thread starter Unassuming
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  • #1
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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][tex]\subseteq[/tex]V for V open, V[tex]\subseteq[/tex]Y.

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

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

Thus, "g o f" is continuous.
 

Answers and Replies

  • #2
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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.
 
Last edited:
  • #3
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It sounds alot better your way. I understand it more now.
 

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