Reading: Read sections 20, 23, 24
Exercises:
Let \(f\) be uniformly continuous on a bounded set \(S\). Prove that \(f\) is bounded (on the set \(S\)).
Proof. Assume that is uniformly continuous on but suppose to the contrary that is unbounded on . Then for all , there exists as number so that . Use these to define the sequence .
is bounded, so by the Bolzano-Weierstrauss theorem, there exists a convergent, hence Cauchy, subsequence . As is assumed to be uniformly continuous, this means that is itself a Cauchy, hence convergent, sequence. Say it converges to the number .
But for all , we have that . As the sequence \((n_k)\) diverges to \(+\infty\), this implies that the sequence \((f_{n_k})\) cannot possibly converge to , a contradiction. Hence must be bounded on .
Give an example of each of the following, or explain why one does not exist.
A function that is continuous on \([a,b]\) but not uniformly continuous.
A function that is continuous on \((a,b)\) but not bounded.
Two functions \(f\) and \(g\) so that neither \(f\) nor \(g\) is continuous at \(x_0 = 1\) but \(fg\) is.
Cannot exist; by theorem 19.2 all functions continuous on \([a,b]\) are uniformly continuous on \([a,b]\).
Consider for instance \(f(x) = 1/x\) on \((0,1)\).
Consider for instance \(f(x) = 1\) except \(f(1) = 0\) and \(g(x) = 0\) except \(g(1) = 10\). Then \((fg)(x) = 0\) which is constant and hence continuous, while neither \(f\) nor \(g\) were at \(x=1\).
Show that \(x = \cos x\) for some \(x \in [0, \pi]\).
Consider \(g(x) = \cos x - x\); \(g(0) = 1\) and \(g(\pi) = 1-\pi < 0\). \(g(x)\) is continuous, so by the Intermediate Value Theorem there exists \(c \in [0, \pi]\) with \(g(c) = 0\), i.e., \(\cos c = c\).
Recall the definition of open set from homework 6.
If \(A\) and \(B\) are open sets, prove that \(A \cap B\) and \(A \cup B\) are open.
Prove that if \((A_i)\) is a sequence of open sets then \(B = \bigcup_{i=1}^{\infty} A_i = \{ x : x \in A_i \textrm{ for some } i \in \mathbb N \}\) is also open.
Find a sequence of open sets \((A_i)\) for which the set \(\bigcap_{i=1}^\infty A_i = \{x : x \in A_i \textrm{ for all } i \in \mathbb N\}\) is not open.
If \(x \in A \cap B\), then \(x\) is interior to \(A\) and \(B\), meaning there exists \(r > 0\) with \(B_r(x) = (x-r, x+r)\) a subset of both \(A\) and \(B\). This means that \(B_r(x) \subseteq A \cap B\). So \(x\) is interior; \(x\) is arbitrary, so the set is open.
If \(x \in A \cup B\), then without loss of generality \(x\) is interior to \(A\), meaning there exists \(r > 0\) with \(B_r(x) = (x-r, x+r)\) which is a subset of \(A\). This means that \(B_r(x) \subseteq A \cup B\). So \(x\) is interior; \(x\) is arbitrary, so the set is open.
If \(x \in \bigcup_{i=1}^\infty A_i\), then \(x\) is interior to \(A_k\) for some \(k\), meaning there exists \(r > 0\) with \(B_r(x) = (x-r, x+r)\) a subset of \(A_k\). So every \(x’ \in B_r(x)\) is in \(A_k\) and hence also in \(\bigcup_{i=1}^\infty A_i\). This means that \(B_r(x) \subseteq \bigcup_{i=1}^\infty A_i\). So \(x\) is interior; \(x\) is arbitrary, so the set is open.
Consider for instance \(A_i = (-1/i, 1/i)\). Then the only element of \(\bigcap_{i=1}^\infty A_i\) is \(0\) (all other reals are not in \(A_k\) for some \(k\) sufficiently large). So \(\bigcap_{i=1}^\infty A_i = \{0\}\), which is not open by definition.