1- Every semi-linear uniform space (X ,Γ) induce a T₁ topological space (X,τ_{Γ}),whose base is B={B(x,V):x∈X,V∈Γ}. More precisely, the topology
τ_{Γ}={G⊆X: for every x∈G there is a V∈Γ such that B(x,V)⊆G}
2- The topology on X×X is the Tychonoff topology induced by Γ on X×X.
Also I think yes, but we need to prove or an idea of the proof.
Taken literally: no. The *family* of members of Gamma is not a subset of XxX, so the question is actually meaningless.
If you ask whether the members of Gamma should be open the the answer is no again: take the normal metric uniformity of the real line; every set that contains a strip of the form {(x,y):|x-y|
Dear Prof. Klaas Hart. I want to clarify three points
1- Each member of Gamma is a subset of XxX, and satisfies the additional assumption, Symmetric and contained the diagonal.
2-If you take the normal metric uniformity of the real line; every set that contains a strip of the form {(x,y):|x-y|
OK, I thought that every uniform space would fall in this class. The family of all closed strips (of positive width) in the plane forms a semi-linear uniformity. The topology it generates, using 1, is the normal topology of the reals. None of the strips is open in the product topology..
Yes this is a semi-linear uniform space. But in this case B(x,V)={y∈X: (x,y)∈V} is the closed interval so the induced topology is the discreet topology. So every strip is open.
OK, since the strips are closed then B(x,V) is a closed interval and every singleton is an intersection of two closed intervals.
Note that I used the `more precisely' version of the definition: "G is open if for every x in G there is V in Gamma such that B(x,V) is in G"; this example shows that the two methods in your definition are not equivalent and in fact, that the family of all B(x,V) needs not be a base for a topology.
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