continuous function

kakarukeys

Is it true that [tex]f[/tex] is a continuous function from [tex]A \times B[/tex] to [tex]C[/tex]
(A, B, C are topological spaces)
if and only if [tex]f_{a}: \{a\}\times B \longrightarrow C[/tex] and [tex]f_{b}: A\times \{b\} \longrightarrow C[/tex] are continuous for all [tex]a\in A, b\in B[/tex] ?
[tex]f_a(b) = f_b(a) = f(a,b)[/tex]

matt grime

The instant reaction to a statement like that in my mind is 'no'.

Let us go for an extreme counter example. Suppose that C has the trivial topology so that only C and the empty set are open. Then any map from AxB to C is continuous as the inverse image of an open set is either empty or AxB. But your f_a is such that the inverse image of C is {a}xB which is rarely going to be open in the product topolgy.

jostpuur

[itex]\{a\}\times B[/itex] is not open in [itex]A\times B[/itex], but it is open in [itex]\{a\}\times B[/itex], which is enough for the continuity.

To me it seems quite clear, that if [itex]f[/itex] is continuous, then also [itex]f_a[/itex] and [itex]f_b[/itex] are continuous. But I'm not sure how it goes in the other direction.

kakarukeys

yes, a restriction is always continuous under the subspace topology.
And the subspace topology is actually the topology of B (or A).
So If f is continuous then f is "continuous in the first variable and second variable".
But is it true if f is "continuous in the first variable and second variable" then f is continuous?

matt grime

No, I doubt it, even modulo my silly error above.

It should be easy to construct counter examples.

The idea ought to be that you cannot write an arbitrary open subset of AxB as a union of things of the form {a}xS and Tx{b} where they are inverse images under the f_a and f_b resp.

At least that is what it feels like without the aid of pencil and paper.

morphism

There's a counterexample in Munkres:

[tex]F : \mathbb{R} \times \mathbb{R} \to \mathbb{R}[/tex]

given by

[tex]
F(x) = \begin{cases}
xy/(x^2 + y^2) &\text{if } (x,y) \neq (0,0) \\
0 &\text{if } (x,y) = (0,0)
\end{cases}
[/tex]

mathwonk

for that one, the restrictions along every horizontal and vertical line are even infinitely differentiable. the example is extremely classical, appearing in Courant, as might be expected, but already in Goursat, page 12 of the 1904 translation by Hedrick. He comments there that such functions were studied by Baire in his thesis.

In particular Baire considered the more difficult question of whether a function can be continuus separately in each variable and yet discontinuous everywhere, as a function of both variables. notice the example above is discontinuous at only one point. He showed I beleieve that such a function is continuous at some points on every vertical, horizontal, and perhaps even every line in its domain.

morphism

Baire... That man was a champion.

mathwonk

Yes but his academic life was a little frustrating, as can easily happen. He was not treated as he deserved and had no approprite position. Lebesgue was preferred to him, in a way he thought unfair,a s Lebesgue had built upon Baire's ideas.

This is just to remind young people that recognition does not always follow achievement.