Looking for a Theorem of Continuous Functions

[sammycaps]

Say I have a function F(x,y)=(f(x),g(y)), F:X×Y→X'×Y'. Is there a theorem that says if f:X→X' and g:Y→Y' are continuous then F(x,y) is continuous. I've proved it, or at least I think I have, but I'd like to know for sure whether or not I'm right.

I know that its not necessarily true that a function defined on a product space is continuous even if it is continuous in each variable separately. But it seems as though since the function I defined above does not interact x and y, there may be some different rules.

Also, if anyone knows for sure that this is not true, that would be useful information as well.

Thanks.

 

[Vargo]

You are right F is continuous.

Suppose U',V' are open in X',Y'. Let U and V be their respective inverse images under f and g. The inverse image of U'xV' by your map F is UxV, which is open in XxY. Then use the fact that all open sets in X'xY' are unions of sets of the form U'xV'.

 

[quasar987]

A function X-->Y x Z, F(x) = (f(x), g(x)) is continuous iff both f:X-->Y and g:X-->Z are continuous.

Your statement is a particular case of this. Also, it would still be true if you had F(x,y)=(f(x,y),g(x,y)).

 

[sammycaps]

A function X-->Y x Z, F(x) = (f(x), g(x)) is continuous iff both f:X-->Y and g:X-->Z are continuous.

Your statement is a particular case of this. Also, it would still be true if you had F(x,y)=(f(x,y),g(x,y)).

Ah I actually was looking at this theorem but I stupidly did not see this. I guess in my case where I define f on X and g on Y, I would write F(x,y)=(f'(x,y),g'(x,y)) where f'(x,y)=f(x)and g'(x,y)=g(x) so then g' and f' are both continuous and defined on X×Y, right?

 

[mathwonk]

as quasar says this follows immediately from what is essentially the defining property of a product. It is also one aspect of the statement that products are functors. i.e. when learning to define a product of two spaces one should also learn to define the product of two maps. i.e. products are functors from the collection (category) of pairs of spaces (X,Y) and pairs of maps (f,g), to the collection (category) of single spaces XxY, and single maps fxg. Your map is fxg.

so the full construction takes a pair of spaces (X,Y) to a space XxY, and a pair of continuous maps (f,g) to a continuous map fxg.

 

[sammycaps]

Can you give me a reference for products of two maps? Wikipedia is weak on this point and google is proving relatively useless.

 

[quasar987]

Ah I actually was looking at this theorem but I stupidly did not see this. I guess in my case where I define f on X and g on Y, I would write F(x,y)=(f'(x,y),g'(x,y)) where f'(x,y)=f(x)and g'(x,y)=g(x) so then g' and f' are both continuous and defined on X×Y, right?

Right.

 

[mathwonk]

you just defined it yourself.

 

[lavinia]

When first learning about continuity I found it instructive to try the theorem first for metric spaces. In this case, I start with X and Y metric spaces and try to come up with a metric on the product space such that Cauchy sequences in the product converge if an only if their projections converge. Then check to see if this gives you the product topology - just to be sure, The more general idea that has been discussed in this thread then becomes obvious from this.

 

[sammycaps]

you just defined it yourself.

Are the components of the product viewed as functions from the product space or from the original space (in the definition I gave I had F(x,y)=(f'(x,y),g'(x,y)) where f(x,y)=f(x) (and for g), but wikipedia just calls them functions from the original spaces)?

 

[sammycaps]

When first learning about continuity I found it instructive to try the theorem first for metric spaces. In this case, I start with X and Y metric spaces and try to come up with a metric on the product space such that Cauchy sequences in the product converge if an only if their projections converge. Then check to see if this gives you the product topology - just to be sure, The more general idea that has been discussed in this thread then becomes obvious from this.

Ok, thanks very much.