Complete Uniform Spaces
November 29, 2007  Posted by John Armstrong  PointSet Topology, Topology
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[...] Miscellany Well, yesterday was given over to examwriting, so today I’ll pick up a few scraps I mentioned in passing on Thursday. [...]
Pingback by Miscellany « The Unapologetic Mathematician  December 1, 2007 
[...] Order on the Real Numbers We’ve defined the real numbers as a topological field by completing the rational numbers as a uniform space, and then extending the field operations to the new points [...]
Pingback by The Order on the Real Numbers « The Unapologetic Mathematician  December 5, 2007 
[...] functions are defined they agree. Since the rationals are dense in the reals (the latter being the uniform completion of the former) there can be only one continuous extension of to the whole real line. We’ll [...]
Pingback by Exponentials and Powers « The Unapologetic Mathematician  April 11, 2008 
[...] We defined the real numbers to be a complete uniform space, meaning that limits of sequences are convergent if and only if they are Cauchy. Let’s write [...]
Pingback by Cauchy’s Condition « The Unapologetic Mathematician  April 23, 2008 
[...] Now we’ve gone through a lot of work to just add one little extra element to our field, but it turns out this is all we need. Luckily enough, the complex numbers are already algebraically complete! This is very much not the case if we were to try to algebraically complete other fields (like the rational numbers). Unfortunately, the proof really is essentially analytic. It seems to be a completely algebraic statement, but remember all the messy analysis and topology that went into defining the real numbers. [...]
Pingback by The Complex Numbers « The Unapologetic Mathematician  August 7, 2008 
[...] we had sequences of rational numbers which didn’t converge to a rational number. Then we completed the topology to give us the real numbers. Well here we’re just doing the same thing! It turns out that the [...]
Pingback by Power Series « The Unapologetic Mathematician  August 18, 2008 
[...] integral does not exist. Yes, is countable, and so it has measure zero. However, it’s also dense, which means is discontinuous everywhere in the unit [...]
Pingback by Iterated Integrals III « The Unapologetic Mathematician  December 18, 2009 
[...] a lemma: if is irrational, then the set of all numbers of the form with and any integers is dense in the real line. That is, every open interval contains at least one point of . The same is true [...]
Pingback by NonLebesgue Measurable Sets « The Unapologetic Mathematician  April 24, 2010 
[...] got a normed vector space, it’s a natural question to ask whether or not this vector space is complete or not. That is, we have all the pieces in place to define Cauchy sequences in our vector space, [...]
Pingback by Topological Vector Spaces, Normed Vector Spaces, and Banach Spaces « The Unapologetic Mathematician  May 12, 2010 
[...] the real numbers form a complete metric space, being Cauchy and being convergent are equivalent — a sequence of finite real numbers is [...]
Pingback by Convergence Almost Everywhere « The Unapologetic Mathematician  May 14, 2010 
[...] We won’t talk quite yet about convergence because our situation is sort of like the one with rational numbers; we have a sense of when functions are getting close to each other, but most of these mean Cauchy [...]
Pingback by The L¹ Norm « The Unapologetic Mathematician  May 28, 2010 
[...] We have a notion of “completeness” for boolean rings, which is related to the one for uniform spaces (and metric spaces), but which isn’t quite the same thing. We say that a Boolean ring is [...]
Pingback by Completeness of Boolean Rings « The Unapologetic Mathematician  August 10, 2010 
[...] old, awkward approach and has nothing to do with category theory — which tells us that every complete metric space is a Baire space. Let be a countable collection of open dense subsets of . We will show that [...]
Pingback by Baire Sets « The Unapologetic Mathematician  August 13, 2010 