Test 2 Proofs Flashcards
Characterization of Compactness in R
A set K ⊆ R is compact if and only if it is closed and bounded
Let K be compact. We will first prove that K must be bounded, so assume, for contradiction, that K is not a bounded set. The idea is to produce a sequence in K that marches off to infinity in such a way that it cannot have a convergent subsequence as the definition of compact requires. To do this, notice that because K is not bounded there must exist an element x1 ∈ K satisfying |x1| > 1. Likewise, there must exist x2 ∈ K with |x2| > 2, and in general, given any n ∈ N, we can produce xn ∈ K such that |xn| > n.
Now, because K is assumed to be compact, (xn) should have a convergent subsequence (xnk). But the elements of the subsequence must satisfy |xnk| > nk, and consequently (xnk) is unbounded. Because convergent sequences are bounded (Theorem 2.3.2), we have a contradiction. Thus, K must at least be a bounded set.
Next, we will show that K is also closed. To see that K contains its limit points, we let x = lim xn, where (xn) is contained in K and argue that x must be in K as well. By Definition 3.3.1, the sequence (xn) has a convergent subsequence (xnk), and by Theorem 2.5.2, we know (xnk) converges to the same limit x. Finally, Definition 3.3.1 (A set K ⊆ R is compact if every sequence in K has a subsequence that converges to a limit that is also in K) requires that x ∈ K. This proves that K is closed
Baire’s Theorem
The set of real numbers R cannot be
written as the countable union of nowhere-dense sets
To start, assume that E1, E2, E3, . . . are each nowhere-dense and satisfy R = infinite union of En
By the definition of En being nowhere-dense, the closure En contains no nonempty open intervals meaning we can apply Exercise 3.5.5 to conclude that the infinite union of En-closure is not equal to R
Since each En is a subset of En-closure, then we know that the infinite untion of En is also not equal to R, which is a contradiction
Uniform Continuity of Compact Sets
A function that is continuous on a compact set K is uniformly continuous on K.
Assume f : K → R is continuous at every point of a compact set K ⊆ R.
To prove that f is uniformly continuous on K we argue by contradiction.
By the criterion in Theorem 4.4.5, if f is not uniformly continuous on K, then there exist two sequences (xn) and (yn) in K such that lim |xn − yn| = 0 while |f(xn) − f(yn)| ≥ ε0 for some particular
ε0 > 0. Because K is compact, the sequence (xn) has a convergent subsequence (xnk) with x = lim xnk also in K.
We could use the compactness of K again to produce a convergent subsequence of (yn), but notice what happens when we consider the particular subsequence (ynk) consisting of those terms in (yn) that correspond to the terms in the convergent subsequence (xnk). By the Algebraic Limit Theorem,
lim(ynk) = lim((ynk − xnk) + xnk) = 0 + x.
The conclusion is that both (xnk) and (ynk) converge to x ∈ K. Because f isassumed to be continuous at x, we have lim f(xnk) = f(x) and lim f(ynk) = f(x), which implies lim(f(xnk) − f(ynk)) = 0.
A contradiction arises when we recall that (xn) and (yn) were chosen to satisfy
|f(xn) − f(yn)| ≥ ε0
for all n ∈ N. We conclude, then, that f is indeed uniformly continuous on K
