# prove that every continuous function is separately continuous

• Nov 21st 2011, 03:02 PM
wopashui
prove that every continuous function is separately continuous
Let (M,d),(N,p),(Y,q) be metric spaces. A function f:MxN-->Y is called separately continuous if for every $\displaystyle a\in M$ and $\displaystyle b \in M$, the function $\displaystyle g:N-->Y$ and $\displaystyle h:M-->Y$, given by $\displaystyle g(y)=f(a,y)$ and $\displaystyle h(x)=f(x,b)$, are continuous.

Prove that every continuous function f:MxN-->Y is separately continuous.

the deinition given is confusing, I have not done any separately continuous question before, need some help here.
• Nov 21st 2011, 07:34 PM
Drexel28
Re: prove that every continuous function is separately continuous
Quote:

Originally Posted by wopashui
Let (M,d),(N,p),(Y,q) be metric spaces. A function f:MxN-->Y is called separately continuous if for every $\displaystyle a\in M$ and $\displaystyle b \in M$, the function $\displaystyle g:N-->Y$ and $\displaystyle h:M-->Y$, given by $\displaystyle g(y)=f(a,y)$ and $\displaystyle h(x)=f(x,b)$, are continuous.

Prove that every continuous function f:MxN-->Y is separately continuous.

the deinition given is confusing, I have not done any separately continuous question before, need some help here.

Prove that you have the obvious embedding $\displaystyle \iota:N\to M\times N:h\mapsto (a,y)$ is continuous and note then that $\displaystyle h=f\circ\iota$, etc.
• Nov 21st 2011, 09:40 PM
wopashui
Re: prove that every continuous function is separately continuous
Quote:

Originally Posted by Drexel28
Prove that you have the obvious embedding $\displaystyle \iota:N\to M\times N:h\mapsto (a,y)$ is continuous and note then that $\displaystyle h=f\circ\iota$, etc.

sorry, what is embedding? can you explain a bit more about the proof?
• Nov 22nd 2011, 06:42 AM
Drexel28
Re: prove that every continuous function is separately continuous
Quote:

Originally Posted by wopashui
sorry, what is embedding? can you explain a bit more about the proof?

There are two ways you can think about it. Perhaps the easiest is, despite what I said, to merely note that $\displaystyle h$ is just $\displaystyle f$ restricted to $\displaystyle \{a\}\times N$ and restrictions of continuous functions are continuous.
• Nov 22nd 2011, 05:29 PM
wopashui
Re: prove that every continuous function is separately continuous
Quote:

Originally Posted by Drexel28
There are two ways you can think about it. Perhaps the easiest is, despite what I said, to merely note that $\displaystyle h$ is just $\displaystyle f$ restricted to $\displaystyle \{a\}\times N$ and restrictions of continuous functions are continuous.

hmm, what about g(x), is g the same as f restricted to Mx{b}?
• Nov 22nd 2011, 07:36 PM
Drexel28
Re: prove that every continuous function is separately continuous
Quote:

Originally Posted by wopashui
hmm, what about g(x), is g the same as f restricted to Mx{b}?

Indeed. Of course, technically we are talking about the restrictions with the obvious identifications $\displaystyle M\approx M\times\{b\}$ and $\displaystyle N\approx \{a\}\times M$, of course this identification is via the embedding I previously mentioned.
• Nov 23rd 2011, 03:57 AM
wopashui
Re: prove that every continuous function is separately continuous
ic, but how does this help for the proof?