So in this post I’ll summarize a bunch of theorems whose proofs we had to know for the final exam, along with the general idea of the proof. This doesn’t even include all the stuff we had to know from the assignments, either… too much stuff to memorize >.< going crazy right now

- Uniform limit of continuous functions is continuous.
**Proof**: Use uniform convergence to find appropriate which has distance less than away from the limit. Find from the continuity of . Apply triangle inequality trick with this particular . - is complete.
**Proof**: Pick any Cauchy sequence. Examine the pointwise sequences; these are Cauchy in (by completeness) and hence converge. Define to be the pointwise limit. Examine as to prove that indeed, the convergence is uniform. Hence the limit is continuous, and a simple argument also shows it’s bounded, so it lives in . - [Weierstrass M-Test] A normed linear space is complete if and only if whenever is a sequence with , we have converges.
**Proof**: The partial sums of the norm series converge and are hence Cauchy. The triangle inequality then implies that the partial sums of the series itself are Cauchy, so by completeness, we get convergence. - [Baire Category] If is a complete metric space and is a sequence of open dense sets in , then their intersection is dense.
**Proof**: We prove that any open set in contains an element of . Choose and with . Now choose and and . And so on. Get a sequence with and so Cantor’s Intersection Theorem. - [Banach Contractive Mapping] If is complete and is contractive of index then there is unique with .
**Proof**: For uniqueness, note that if and then means . For existence, just pick an arbitrary point and iterate over and over to get a sequence. Apply triangle inequality to obtain Cauchyness, and hence convergence. Continuity of gives that the limit is a fixed point since and . - [Arzela-Ascoli (one direction)] Let be compact. If then it is relatively compact if and only if it is pointwise bounded and equicontinuous.
**Proof**(forward): Relatively compact implies totally bounded, so get a finite -mesh for . is compact, so continuity is always uniform. Exploit the uniform continuity of the stuff in the mesh, and then exploit the fact that it’s a mesh. Triangle inequality trick: the proof is basically a replica of the “uniform limit of continuous functions” theorem. We get uniform boundedness since the stuff in the mesh is bounded (and it’s a mesh). - [Weierstrass Approximation] If then there is a sequence of polynomials converging uniformly to on .
**Proof**: Can’t really summarize this nicely at the moment. Not exactly my favourite proof -_-. - [Stone-Weierstrass (Lattice Version)] If is compact and we have a point-separating subspace which contains the function 1 and also contains for any two functions then is dense in .
- [Stone-Weierstrass (Subalgebra Version)] If is compact and we have a point-separating unital subalgebra then it is dense in .
**Proof**: Can assume is closed. Invoke Weierstrass Approximation Theorem to find a sequence of polynomials converging uniformly to the absolute value function on where is a bound for . Hence by considering , prove that . Thereafter, prove that and invoke Lattice Version. - is not .
**Proof**: Assume it is. By manipulating some intersections and unions, we quickly derive that is first category, contradicting Baire Category. - is compact if and only if every family of closed sets with FIP has nonempty intersection.
**Proof**: Since each is closed, is open. So if the intersection*is*empty, then by taking its complement, we actually get an open cover of ! Now we can take a finite subcover, complement again and hence contradict FIP. Other direction is equally simple. - Continuous image of compact set is compact.
**Proof**: If we have an open cover of we can pull each set back, and by continuity of the pullbacks are all open, yielding an open cover of . Pick finite subcover, push it the other way and you’re done. - Continuity on a compact set is always uniform.
**Proof**: Let and for all pick . Then write . Pick a finite subcover and then let be the minimum of all those appearing in this subcover. Done. - If is compact then every infinite set has a cluster point.
**Proof**: Just pick a sequence from this set with all the terms distinct. By compactness, it has a convergent subsequence and hence a cluster point, by distinctness. - If is sequentially compact then it is totally bounded.
**Proof**: If we assume it is not totally bounded then there exists an such that no -mesh exists, so we can pick a sequence whose terms are always at least apart. This will (obviously) not have a convergent subsequence, contradicting sequential compactness.

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Wow, I only understood # 1 and # 3. I have a lot of catching up to do in MATH247 before I start PMATH351.

The other proofs do look very interesting though.

should just read my 351 notes. it’s probably all perfectly accessible to you anyway.

+1

Here’s hoping my limited background in multivariate calculus will be enough…

in the 3rd last one “Continuity on a compact set is always uniform” you have an interval where you should have a ball