Upper and Lower Bound
Def: Let \(S \subseteq \mathbb{R}\) and \(r \in \mathbb{R}\). We say that
(1.1) \(r\) is an upper bound of \(S\) : \(\quad \forall s \in S \, [s \le r]\)
(1.2) \(r\) is an lower bound of \(S\) : \(\quad \forall s \in S \, [s \ge r]\)
(2) \(r\) is the greatest element of \(S\) (notation \(\max S\)):
1. \(r\) is an upper bound of \(S\).
2. \(r \in S\)
(3.1) \(r\) is the least upper bound of \(S\) (notation \(\sup S\)): \(\quad r = \min \{\text{ upper bounds of } S\}\)
(3.2) \(r\) is the greatest lower bound of \(S\) (notation \(\inf S\)): \(\quad r = \max \{\text{ lower bounds of } S\}\)
We have
Any \(u < \sup S\), \(u\) is not upper bound of \(S\).
Any \(u > \inf S\), \(u\) is not lower bound of \(S\).
Also, every \(r \in \mathbb{R}\) is a upper and lower bound of \(\emptyset\).
Convention:
We write \(\sup S = \infty\) if and only if \(S\) has no upper bound. If this is the case we say \(\sup S\) doesn’t exist.
We way \(S\) is bounded from above iff \(S\) has a upper bound.
Dedekind Cut
If we split the set \(\mathbb{R}\) to two non-empty set, lower part \(A\) and upper part \(B\). Then either \(\max A\) exists or \(\min B\) exists.
Def (Dedekind Cut): Let \(A, B \subseteq \mathbb{R}\), we say that \((A,B)\) is a Dedekind cut (of \(\mathbb{R}\)) if
\(A \ne \emptyset \ne B\)
\(A \cup B = \mathbb{R}\)
\(\forall a \in A, b \in B, [a < b]\)
We usually call \(A\) (resp. \(B\)) the lower (resp. upper) part of \((A,B)\).
From now on (until he say stop) we assume that \(\mathbb{R}\) has the following property:
(Dedekind’s gapless property) If \((A,B)\) is a D-cut of \(\mathbb{R}\), then exactly one of the following happens:
\(\max A\) exists but \(\min B\) doesn’t.
\(\min B\) exists but \(\max A\) doesn’t
Ex 1: We may define Dedekind cuts of \(\mathbb{Q}\) similarly.
\[B = \{x \in \mathbb{Q} \vert x > 0 \land x^2 > 2\}, A = Q \backslash B\]We can verify it is D-cut.
We can prove \(\max A, \min B\) doesn’t exists.
If \(\min B\) exists, let \(\min B = p\).
\[\min B = p \implies p \in B \implies p > 0 \land p^2 > 2\]Let \(q = p - \dfrac{p^2-2}{2p} \implies 0 < q < p\)
\[q^2 = \Big( p - \dfrac{p^2-2}{2p}\Big)^2 = 2 + \Big(\dfrac{p^2-2}{2p}\Big)^2 > 2\] \[q > 0 \land q^2 > 2 \land q \in \mathbb{Q} \implies q \in B\]contradicts with \(q < p = \min B\)
If \(\max A\) exists, let \(\max A = p\)
\[\max A = p \implies p \in A \implies p^2 < 2\] \[1 \in A \implies p \ge 1 > 0\]Let \(h = \min\Big( \dfrac{1}{2}, \dfrac{2-p^2}{2p+2}\Big)\)
\[h \le \dfrac{1}{2} < 1\] \[h \le \dfrac{2-p^2}{2p+2} < \dfrac{2-p^2}{2p+1}\] \[(p+h)^2 = p^2 + (2p+h)h < p^2 + (2p+1)h < p^2 + (2-p^2) < 2\] \[p + h \in Q \land (p+h)^2 < 2 \implies p+h \in A\]contraditcs with \(p+h > p = \max A\)
Weierstrass
Theorem (Weierstrass): Let \(\emptyset \ne S \subseteq \mathbb{R}\). If \(S\) has a upper bound, then \(\sup S\) exists.
Proof:
\[\sup S = \min \{\text{upper bounds of } S\}\]The idea is to do a Dedekind’s cut \(( \mathbb{R} \backslash \{\text{ upper bounds of } S\} ,\{\text{upper bounds of } S \})\), then we prove the other case is impossible.
Let \(B := \{b \in \mathbb{R} \vert b \text{ is a upper bound of } S\}, A:= \mathbb{R} \backslash B\)
We need to show that \(\min B\) exists.
Step 1: \((A,B)\) is a D-cut of \(\mathbb{R}\)
- \(A \ne \emptyset \ne B\)
\(A \cup B = \mathbb{R}\)
\(\forall a \in A, b \in B, [a < b]\).
Were this false, \(a \ge b\)
\[b \in B \implies a \in B \implies a \in A \cap B \implies a \in \emptyset\]Step 2: We need to show that \(\max A\) doesn’t exists. Were this false.
\[\max A = a_0 \implies a_0 \in A \implies a_0 \text{ is not an upper bound of} S \implies \exists s_0 \in S \text{ s.t. } s_0 > a_0\]Let \(h = \dfrac{a_0 + s_0}{2}, a_0 < h < s_0\)
\[h < s_0 \implies h \text{ is not an upper bound of } S \implies h \in A\]contracts with \(h > a_0 = \max A\)
The Archimedean Property
Ex2: prove the following statement
(The Archimedean property) \(\forall r \in \mathbb{R} [r > 0 \implies \exists n \in N [\frac{1}{n} < r]]\)
Step 1: We need to show \(\mathbb{N}\) doesn’t have upper bound. Were this false, \(\mathbb{N}\) has upper bound, then from Weierstrass \(\sup \mathbb{N}\) exists, \(b = \sup \mathbb{N} = \min \{\text{ uppber bound of } \mathbb{N}\}\).
\[\forall n \in \mathbb{N}, [n+1 \in \mathbb{N}]\] \[n+1 \in \mathbb{N} \implies n + 1 \le b \implies n \le b - 1\] \[\forall n \in \mathbb{N}, [n \le b - 1]\]thus \(b - 1\) is an uppber bound of \(\mathbb{N}\), \(b - 1 \in \{ \text{ upper bound of } \mathbb{N}\}\) contracdicts with
\[b - 1 < b = \min \{ \text{ uppber bound of } \mathbb{N} \}\]Step 2: Were Archimedean property false
\[\exists r \in \mathbb{R} \land r > 0 [\forall n \in \mathbb{N} [\frac{1}{n} \ge r]]\]thus
\[\exists r \in \mathbb{R} \land r > 0 [\forall n \in \mathbb{N} [n \le \frac{1}{r}]]\]thus \(1/r\) is an upper bound of \(\mathbb{N}\)
Limit of Sequence
Def: Let \(a_n (n \in \mathbb{N})\) be a sequence in \(\mathbb{R}\), and \(L \in \mathbb{R}\). We say that \(a_n\) converges to \(L\) (as \(n \to \infty\)) if
\[\forall \varepsilon > 0 \quad \exists N \in \mathbb{N} \quad [n \ge N \implies \vert a_n - L \vert < \varepsilon]\]Terminology: If such \(L\) exists (doesn’t exists), we call it the limit of \(a_n\) and we call \(a_n\) a convergent (divergent) sequence.
the \(n \ge N\) can be freely changed to \(n > N\).
the \(\vert a_n - L \vert < \varepsilon\) can be freely changed to \(\le\) and \(K \varepsilon\), where \(K \in \mathbb{R}\) is a constant not depending on \(\varepsilon, N\).
Notation: \(\lim_{n \to \infty} a_n = L\)
Some generalized notations: \(\lim_{n\to\infty} a_n = \infty \quad \iff \quad\) \(\forall M \in \mathbb{R}\, \exists n \in \mathbb{N} \, [n \ge N \implies a_n \ge M]\), similarly for \(-\infty\). In this case we usually doesn’t say the limit exists or converge.
Ex3: (1) \(\lim_{n \to \infty} a_n = L, \lim_{n \to \infty} a_n = M \quad \implies \quad L = M\)
wlog, assume \(M > L\)
\[\exists N_1 \in \mathbb{N} \, [n \ge N_1 \implies L - \dfrac{M - L}{2} < a_n < L + \dfrac{M-L}{2}]\] \[\exists N_2 \in \mathbb{N} \, [n \ge N_2 \implies M - \dfrac{M - L}{2} < a_n < M + \dfrac{M-L}{2}]\] \[n \ge \max(N_1, N_2) \implies \dfrac{M+L}{2} < a_n < \dfrac{M+L}{2}\]we have the contracdiction.
(2) \(a_n (n \in \mathbb{N})\) is convergent \(\implies \{a_n \vert n \in \mathbb{N}\}\) is bounded.
\[\exists N \in \mathbb{N}, [n \ge N \implies L - 1 < a_n < L + 1]\]let \(L_1 = \max \{a_n \vert n \in \mathbb{N}, n < N\}\), then
\[\forall n \in \mathbb{N} \, [a_n \le \max(L+1, L_1)]\]lower bound is similar.
(3) If \(\forall n \in \mathbb{N} a_n \le b_n, \lim_{n\to\infty}a_n = L, \lim_{n\to\infty}b_n=M\), then \(L \le M\). What if \(\le\) is replaced by $<$?
Were this false, \(L > M\). We can find \(N_a, N_b \in \mathbb{N}\) such that \(a_n, b_n\) all lies in the nonoverlaping region. Then we find \(n \ge \max(N_a, N_b) \implies a_n > b_n\).
If \(\le\) is replaced by \(<\), we still only have the conclusion \(L \le M\). Assume the examlpe \(a_n = \dfrac{1}{n+1}\), \(b_n = \dfrac{1}{n}\). By Archimedean property, limit is \(L=M=0\).
Remark: changing or removing finitely many terms in \(a_n\) doesn’t affect the existance of limit or the value of the limit.
Elementary Arithmetic on Limit
If \(\lim_{n\to\infty} a_n = L\) and \(\lim_{n\to\infty} b_n = M\), then
(1) \(\lim_{n\to\infty} (a_n \pm b_n) = L \pm M\)
(2) \(\lim_{n\to\infty} a_n b_n = L \cdot M\)
(3) if \(M \ne 0\), then \(b_n \ne 0\) for all but finitely many \(n\), and \(\lim_{n\to\infty} \dfrac{a_n}{b_n} = \dfrac{L}{M}\)