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"Any HN'ers tried to hack their way through Baby Rudin?" Yup. Been there; done that! Yes, it needs some 'explanation': Q. Why do we care about metric spaces
by NY_USA_Hacker 15y ago
"Any HN'ers tried to hack their way through Baby Rudin?"
Yup. Been there; done that!
Yes, it needs some 'explanation':
Q. Why do we care about metric spaces?
A. Eventually they are important, e.g., in Hilbert and Banach spaces. For now, they are the vanilla case of a topological space.
Q. Why do we care about topological spaces?
A. Because they are the nicest way to describe continuity. So a function from one topological space to another is 'continuous' iff (from Halmos: if and only if!) for each open set in the range space its inverse image under the function is an open set in the domain space. That's a nicer, and more general, definition than converging sequences (which are not quite as powerful, but examples are a bit bizarre) or epsilon-delta arguments from limits in calculus.
Q. An 'open' set?
A. A 'topology' is defined as a collection of sets it calls 'open'. So, really the topology is defined just by some open sets. A good and easy example of an open set is an open interval, say, (0,1) on the real line. That set is 'open' in the 'usual' topology (in principle there can be others) for the real line.
Q. Why do we care about continuity?
A. As in G. Simmons, the two main pillars of the 'analysis' part of math are linearity and continuity. They are assumptions that commonly hold in practice and that give some of the most powerful and useful results. In particular, in Baby Rudin, if a real valued function of a real variable is continuous on a closed set, then the function is also uniformly continuous, is bounded, actually achieves its upper and lower bounds, and its Riemann integral exists and is finite. All nice stuff.
Continuing, all such functions form a vector space (a 'linear' space); the difference to two such functions is again such a function; and the maximum value of the difference forms a norm. So, those functions form a normed linear space. If a sequence of such functions converges in this norm, then the limit is also such a function -- i.e., a uniform limit of continuous functions is continuous. So, the space is 'complete' in this norm. So, we have the leading example of a Banach space (a complete, normed linear space). Then we can show the Hahn-Banach result and see the tasty, ice cream and cake dessert examples in Luenberger, 'Optimization by Vector Space Methods' including Kalman filtering, results for 'least action' in physics, more in deterministic optimal control, etc. So, these are some of the coming attractions of those first chapters in Baby Rudin on metric spaces, topology, open sets, continuity, etc.
Q. A 'closed' set?
A. Back to the real line where (0,1) is an open set, [0,1] is a closed set, So, it's closed and bounded. Presto: It is the leading example of a 'compact' set, which is almost as nice to work with as a finite set. In particular, any infinite subset of a compact set has a limit point in the set which is the key to uniform continuity, the existence of the Riemann integral, and more. The nicest definition of a compact set is that each open cover has a finite subcover, and this definition works also in metric spaces and topological spaces.
So here is an explanation of what the first chapters of Baby Rudin are all about. Print a copy and pin it to the wall in any math department that teaches a Baby Rudin course!
- NY_USA_Hacker 15y agoErrata: Of course, in "In particular, in Baby Rudin, if a real valued function of a real variable is continuous on a closed set, then the function is also uniformly continuous, is bounded, actually achieves its upper and lower bounds, and its Riemann integral exists and is finite." replace "on a closed set" with "on a closed and bounded set".