Topology of real number space

Def'n. Cauchy sequence

\[\forall \epsilon > 0. \exists N >0 .\forall m,n \geq N. |a_m - a_n| \leq \epsilon\]

Any convergent sequence is Cauchy
Every Cauchy sequence is bounded

Def'n. Metric space

a space X plus a notion of distance

Def'n. Completeness

a metric space \(S\) is complete if Cauchy sequence in \(S\) converges to some point in that space.

Claim (D&D 2.8.5)

every Cauchy sequence of reals converges. i.e. \(\mathbb{R}\) is complete.

proof. Let \((x_n)\in\mathbb{R}\) be Cauchy and \(x_n\rightarrow L\). WTS \(L\in\mathbb{R}\)

We'll use diagonalization argument: For fixed \(x_n\in\mathbb{R}, \exists (r_{n,k})\) be rational Cauchy s.t. \(\lim_{k\rightarrow\infty}r_{n,k}=x_n\). Let \(a_n\) be s.t. \(|a_n - x_n|\leq 1/n\) by choosing a candidate for each \(n\).

\[\implies |a_n-L|\leq |a_n-x_n| + |x_n-L|\leq 1/n + \epsilon_n \implies \lim_{n\rightarrow\infty} a_n = L \in\mathbb{R}\]

Thrm. Mean Value Theorem

\[\exists \xi\in(x,y).|f(x)-f(y)| = |f'(\xi)||x-y| \]

Example

\(a_1 = 1, a_{n+1} = cos(a_n)\) show that \(a_n\rightarrow L\).
proof. WTS \((a_n)\) is Cauchy. Take some \(\xi\in(a_n,a_{n-1})\), then

\[\begin{align*} |\cos(a_{n - 1})-\cos(a_n)|&\leq |\sin(\xi)||a_{n-1}-a_n|\\ \implies |a_n - a_{n+1}|&\leq r|a_{n-1}-a_n|\\ &\leq r^n|a_2 - a_1| \end{align*}\]

If \(r < 1\), then \(\lim|a_n - a_{n-1}|=0\)

Close and Open

Limit point For a sequence \((x_n)\in\mathbb{R}\), its limit point \(x\in\mathbb{R}\) satisfies \(\lim_{n\rightarrow\infty}x_n = x\).

Closed if it contains all its limit points

Example \(\mathbb{R}^n\) is closed because \(\mathbb{R}\) is complete

Claim Finite union of closed sets is closed

Open if \(\forall x\in U.\exists r_{x,v}>0. B_r(x)\subseteq U\)

\(U\) open, \(f\) continuous \(\Rightarrow f^{-1}(U)\) is open

Claim (D&D 4.3.8) \(U\) open IFF \(\mathbb{R}-U\rightarrow U^c\) is closed

Theorem

finite intersection of open sets is open.

proof. \((\cap^M U_i)^{c^c} = (\cup^M U_i^c)^c\)
Since \(U_i\) open, \(U_i^c\) closed, \(\cup^M U_i^c\) closed.

Compactness

Compact Defn 1 \(C\subseteq\mathbb{R}\) is compact if all sequence \(\{x_n\}\subseteq C\) has a converging subsequence \(\{x_{n_k}\}\subseteq C\)
\(\forall \{x_n\}\subseteq C. \exists \{x_{n_k}\}\subseteq \{x_n\}. x_{n_k}\rightarrow x\in C\)

Compact Defn 2 A set \(C\) is compact if every open cover of \(C\) has a finite subcover that covers \(C\).

Open cover A collection of open balls \(\{U_a\}_{a\geq 1}\) is an open cover if \(c\subseteq \cup^\infty U_a\)

Claim \([a,b]\subseteq \mathbb{R}\) is compact.

proof. By Bolzano Weierstrass Theorem. every bounded sequence has a convergent subsequence. Since \([a,b]\) closed, such limit falls in \([a,b]\).

Continuous mapping theorem For continous \(f, x_n\rightarrow x\Rightarrow f(x_n)\rightarrow f(x)\).

Corollary \(f:U\rightarrow V\), if \(C\subseteq U\) is compact, then \(f(C)\) is also compact.

Thrm. Heine-Borel Theorem

\(C\subseteq \mathbb{R}\) is compact IFF closed and bounded.

Claim 1 compact \(\Rightarrow\) bounded

proof. Suppose compact but not bounded. Then take \(\{x_n\}\subset C\), wlop, \(x_n\rightarrow \infty\). Then, we can take a monotone increasing subsequence \(y_n\) of \(x_n\). Then by compactness, take \(y_{n_k}\rightarrow y\in C\). But a monotone divergent sequence cannot have convergent subsequence.

Claim 2 compact \(\Rightarrow\) closed

proof. Let \(\{x_n\}\subseteq C. x_n\rightarrow x\), WTS \(x\in C\).
By compactness, take \(\{x_{n_k}\}\rightarrow L\). We can show that \(|x-L| = |x - x_{n_k} + x_{n_k} - L| \leq |x - x_{n_k}| + |x_{n_k} - L| \rightarrow 0\)

Claim 3 closed & bounded \(\Rightarrow\) compact

proof. Let \(\{x_n\}\subseteq C\). Since bounded, by BWT, take \(\{x_{n_k}\}\) where \(x_{n_k}\rightarrow L\in\mathbb{R}\).
Since \(C\) closed, its limit point is in \(C\).

Thrm. Extreme value theorem

If a function \(f:K\rightarrow R\) is continous and \(K\) is compact, then \(f(K)\) is compact and \(\exists a,b\in K. \forall k\in K. f(a)\leq f(k)\leq f(b)\)

Part1 \(f(K)\) compact
proof. Let \((y_k)\subseteq f(K)\), take \((x_k)\subseteq K s.t. f(x_k)=y_k\). By compactness of \(K\), take \((x_{k_n})\subseteq (x_k)\) and \(x_k\rightarrow x\in K.\) Then by continuous mapping theorem, \(f(x_{k_n})\rightarrow f(x)\in f(K)\)

Part2 \(f(K)\) has max and min
proof. By compactness and least bound principle, take \(M,m\) be the supremum and infimum of \(f(X)\). WTS thats contained in \(f(K)\).
Construct a sequence \((d_n)\subset K\) s.t. \(M-\frac{1}{n}\leq f(d_n)\leq M\) (since M is the supremum we can do such construction).
Since \(K\) is compact, by Bolz. Weis. Theorem, take \((d_{n_k})\subseteq (d_n), d_{n_k}\rightarrow d\) is convergent.

Then by squeeze theorem, \(M-\frac{1}{n}\leq f(d_{n_k})\leq M\) is squeezed by \(M\), \(\lim f(d_{n_k}) = f(\lim f(d_{n_k})) = M\). By compactness of \(f(K), M\in f(K)\).

Example \(f\) continuous on \([a,b]\subseteq \mathbb{R}\) has no local extremum, then \(f\) is monotone.
local maxmimum/minimum \(x_0\) is local max/min if \(\exists \epsilon > 0. \forall x\in [x_0-\epsilon, x_0+\epsilon], f(x)\leq | \geq f(x_0)\).

Uniform Continuilty

\(f:S\rightarrow R\) is uniformly continuous if \(\forall \epsilon > 0. \exists \delta > 0. \forall x,y. |x-y|\leq \delta \Rightarrow |f(x)-f(y)|<\epsilon\)

Sequential criterion A function \(f:S\rightarrow R\) is not uniform continuous if \(\exists (x_n), (y_n)\subseteq S\) s.t. \(|x_n - y_n|\rightarrow 0\) but \(\exists \epsilon_0 >0 s.t. |f(x_n)-f(y_n)|\geq \epsilon_0 > 0\)

proof. negation of the uniform continuous definition

Example \(f(x):=x^2\)
Let \(x_n = n + n^{-1}, y_n = n\). Then \(|x_n-y_n|= n^{-1}\rightarrow 0\), but \(|(n+n^{-1})^2 - n^2| = |-2|+n^{-2}\rightarrow 2 > 0\)

\(\alpha\)-Holder \(f:S\rightarrow R\) is a-Holder if \(|f(x)-f(y)|\leq c |x-y|^\alpha, \alpha\in[0,1]. \forall x,y\in S\).

Example \(f(x)=x^p, p\in(0,1)\) is call p-Holder and is uniform continuous.

proof. Let \(\epsilon > 0\), take \(\delta = (\epsilon / c)^{a^{-1}} < 1\) because \(\alpha \leq 1\)
\(|f(x)-f(y)|\leq c|x-y|^\alpha \leq c\delta^\alpha =\epsilon\)

When \(\alpha = 1, i.e. |f(x)-f(y)|\leq c|x-y|\), they are called Lipschitz.

Example Any differentiable functions are Lipschitz (bounded derivative)

proof. by MVT, \(|f(x)-f(y)| = |f'(c)||x-y|\leq B|x-y|\)

Thrm. Heine-Cantar Theorem

if \(K\) compact, and \(f:K\rightarrow R\) is continuous, then \(f\) is also uniform continuous.

proof. Suppose \(f\) is not uniform continuous
Take \((x_n), (y_n)\) and for \(\epsilon > 0\) s.t. \(|x_n - y_n|\leq 1/n\) but \(|f(x_n)-f(y_n)|\geq \epsilon > 0\). By \(K\) compact, take \((x_{n_k})\rightarrow x\in K\)。
However, \(|y_{n_k}-x|\leq |y_{n_k}-x_{n_k}| + |x_{n_k}-x| \leq n_k^{-1} + \epsilon_k \rightarrow 0\) i.e. \(y_{n_k}\rightarrow x\)
\(\forall n_k > N_{\epsilon_0/4}|f(x_n)-f(y_n)|\leq |f(x_n)-f(x)| + |f(x)-f(y)| \leq \epsilon_0/2\) causes contradiction

Thrm. Intermediate value theorem

If \(f:[a,b]\rightarrow \mathbb{R}\) is continuous and \(z\in\mathbb{R}\) satisfy \(f(a)\leq z\leq f(b)\), then \(\exists c \in [a,b]\) s.t. \(f(c)=z\).

proof. Let \(A = \{x\in[a,b]: f(x) < z\}\), then \(A\) is upper bounded. Then, take \(c = sup(A)\).

\(\forall \epsilon_n = n^{-1}\), take \(a_n \in A\) s.t. \(c - n^{-1} \leq a_n \leq c\) and \(f(a_n)\leq z\) so \(f(c) = \lim_{n\rightarrow \infty} f(a_n)\leq z\) (1).
Explicitly assume \(c < b\) (otherwise we've done), then take \(\exists b_n\) s.t. \(c < b_n \leq b, \lim_\infty b_n =c\).
Since \(z\) is upper bound for \(A, b_n\not\in A, f(b_n)\leq z\) (2). By (1)(2) \(f(c)=z\)

Connected

Disconnected \(A\subseteq \mathbb{R}\) is disconnected \(U,V\) open cover s.t. \(A = U\sqcup V\)
Connected \(A\) is connected if \(\forall U,V. A\neq U\sqcup V\)
Path connected of \(\forall a,b \in A. \exists \gamma:[0,1]\rightarrow A\) be a continuous path i.e. \(\gamma(0)=a, \gamma(1)=b, \gamma([0,1])\subseteq A\)