Showing that a given function is continous over a certain topology

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The discussion centers on the continuity of the functions h: R^ω → R^ω defined by h(x1, x2, x3,...) = (x1, 4x2, 9x3,...) and g(t) = (t, t, t, t, t,...), specifically under the product topology, box topology, and uniform topology. The user struggles with applying the definition of continuity, which requires that the preimage of an open set must also be open. The example provided for g illustrates that it is not continuous in the box topology, as the preimage of an open set results in the closed singleton {0}. This highlights the failure of the box topology in certain continuity tests.

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Consider the maps h: R^w (omega) ---> R^w (omega) , h (x1, x2, x3,...) = (x1,4x2, 9x3,...)
g: (same dimension mapping) , g (t) = (t, t, t, t, t,...)

Is h continuous whn given the product topology, box topology, uniform topology?

For the life of me i am having trouble trying to understand what i need to do to accomplish this. I know that the definition of continuous in topology is that the pre image f^-1 (v) of an open set v has to be open, but how do i apply that to this sort of situation?

Take the function h, am i assuming that (x1, 4x2, 9x3,...) is my open set or am i going to take an arbitrary basis element of my open set that looks like (x1, 4x2, 9x3,...) apply the preimage to that, and then if it is open then i can conclude the function is continuous?

I saw a solution for the function g, and they said it is not open in the box topology. They took th open set [-1/(n^2) , 1/(n^2)] and said taking the inverse image of g the set is {0}. How did they do this? I fail to see the connection. Please help I've been stuck here all day tryimg to make sense of this.

 
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I imagine that for g, they were taking the limit as ##n \to \infty##. This would give you an open set for the image, but in the limit, the pre-image is the closed singleton {0}.
 
RUber said:
I imagine that for g, they were taking the limit as ##n \to \infty##. This would give you an open set for the image, but in the limit, the pre-image is the closed singleton {0}.
Yes, but how do they get that singleton is my question. Is it that you look at each inverse image of each coordinate function individually and take the intersection of those , which produces the common point 0?
 
Assume that ##U## is an open set ##U=(-1/n^2, 1/n^2)##, for ##V=g^{-1}(U) ## to be open, there must be ##\epsilon \in V, \, \epsilon \neq 0 ##, such that ##g(\epsilon) \in U##. But for any epsilon, there is an ##n ## that makes ##g(\epsilon) \notin U##.
You could also think of ##U = (-\epsilon, \epsilon)## if this helps.
Wikipedia uses exactly this example to describe how box topology fails at continuity tests: http://en.wikipedia.org/wiki/Box_topology
 

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