I am trying to do excercise 4.1. in Continuous Time Markov Processes: An Introduction by Thomas Ligett. There I have to provide some counterexample of a function in C({0,1}N). However I dont

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intacte87 · Accepted Answer

{0,1}N has the product topology. A sequence converges in this space iff each coordinate converges. Since the coordinates are just 0 or 1 this means each coordinate becomes constant after some stage. An example of dis-continuous function on this space is given by  f((xn))=∈f{n:xn=1} if xn=1 for some n and f(0,0,,…)=0. Note that f(en)=n if en is the sequence qwith 1 in the n-th place and 0 elsewhere. Also, en→(0,0,…). Clearly f(en) does not tend to f(0,0,…) so f is not continuous.

Chanell Sanborn · Accepted Answer

Your original question is about the definition of continuity in your topological space. (You, yourself, stated a valid metrization. That is a good way of understanding a topology for a lot of mathematicians working in probability theory.) It sounds like what you might want is an example for the problem you mention later. Here is an example that might be somewhat canonical:   What about f(η)=lim ⊃n→∞fn(η) where we define fn(η)=n−1∑i=1nη(i)? This is a stand-in for the asymptotic density of η. Then f(ηx)=f(η) (where ηx is the configuration obtained by changing η exactly at the site x) because for any given n&amp;gt;x we have |fn(η)−fn(ηx)|=n−1 and limn→∞n−1=0. So we definetely have ∑x∈N⊃η∈{0,1}N|f(ηx)−f(η)| equals 0 because it is the series of all 0s.

alenahelenash · Accepted Answer

You can prove continuity using the metric and ϵ−δ argument, but this seems to be unnecessary. Just use the definition of the product topology, show that if x=(x1,x2,⋯)∈f−1(U) where U is an open set of R, then there exists n such that (x1,x2,⋯,xn,y1,y2,⋯)∈f−1(U) for all (y1,y2,⋯)∈{0,1}N.

I am trying to do excercise 4.1. in Continuous Time Markov Processes: An Introduction by Thomas Lige

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