Real Analysis
I am finding Real Analysis to be more difficult than any other mathematics that I have studied before. I can seem to verify the truth of statements because they seem right; but I am having a difficult time producing rigorous and correct proofs.
It seems that High-School children (on the internet) are able to self-study Fomin with success. Bitterly, we remind ourselves:
“Comparison is the thief of Joy”—Theodore Roosevelt (probably)
Exercises
Results
Set Theory
A Cartesian product of non-empty sets is non-empty.
\(f: A\rightarrow B\) \[f \subseteq A \times B \iff \forall x \in A,\; \exists !y \in B | (x,y) \in f\]
\(f: A\hookrightarrow B\) \[\forall x_1, x_2 \in A,\; f(x_1) = f(x_2) \implies x_1 = x_2 \]
\(f: A\twoheadrightarrow B\) \[\forall y \in B, \exists x \in A \mid f(x) = y\]
(injective and surjective) \[\forall y \in B,\; \exists !x \in A \mid f(x) = y\]
We say that two sets \(A\) and \(B\) have the same cardinality if there is a bijection \(f: A\rightarrow B\); we then write \(A\sim B\), which is the same as \(|A| = |B|\).
Also, if there is an injective function \(f:A\rightarrow B\), we say \(|A|\leq |B|\). Or, equivalently, a surjection from \(f: B\rightarrow A\).
Let \(S\) be any set, and let \(\mathcal{P}(S)\) be its power set. Then \(S \not\sim \mathcal{P}(S)\).
Let \(A\) and \(B\) be sets, and suppose that there exist injective functions \(f:A\rightarrow B\) and \(g: B\rightarrow A\). Then there exists a bijective function \(h:A\rightarrow B\).
A set \(S\) is finite \(|S| = {1,\ldots,n}\), for some \(n\in\mathbb{N}\). Otherwise \(S\) is infinite.
A set \(S\) is Dedekind-infinite if there is a bijection from \(S\) to a proper subset of itself. Otherwise \(S\) is Dedekind-finite.
We say that a set \(S\) is countable if \(|S| \leq |\mathbb{N}|\). Otherwise \(S\) is uncountable. If \(S\) is countable and infinite we say that \(S\) is countably infinite.
Metric Spaces
\(\displaystyle\lim_{x \rightarrow a} f(x) = b\) means that “for any number \(\varepsilon > 0\), there is a number \(\delta(\varepsilon)\) such that \(|f(x)-b| < \varepsilon\) whenever \(|x-a|<\delta\)”
A metric space is a pair \((X,d)\), where \(X\) is a (non-empty) set and \(d: X \times X \rightarrow [0,\infty)\) is a function, such that the following conditions hold for all \(x,y,z \in X\):
- \(d(x,y) = 0 \iff x=y\)
- \(d(x,y) = d(y,x)\)
- \(d(x,y) + d(y,z) \geq d(x,z)\quad\) (triangle inequality)
A sequence in a set \(X\) is a function from \(\mathbb{Z}^+\) to \(X\). \[\set{x_n}^\infty_{n=0}\]
\(\set{x_n}^\infty_{n=0} \subset \mathbb{R}\) converges to a limit \(x\in \mathbb{R}\) if for every \(\epsilon > 0\), there is a \(K(\epsilon)\in \mathbb{N}\) such that $|x_n - x| < ε $ whenever \(n > K(\epsilon)\).
A sequence in a metric space can have at most one limit
A function \(f: \mathbb{R} \rightarrow \mathbb{R}\) is continuous if for every \(x\in \mathbb{R}\) and every \(\epsilon > 0\), there is a \(\delta(x, \epsilon) > 0\) such that \(|f(y) - f(x)| < \epsilon\) whenever \(|y-x|<\delta(x,\epsilon)\).
For a point \(x\) in a metric space \((X,d)\) and a number \(\epsilon > 0\), define the $ε$-ball \[B(x,\epsilon) = \set{y\in X: d(y,x) < \epsilon}\]
Let \((X,d)\) be a metric space, and consider \(Y\subseteq X\). Define the interior \[\mathrm{Int}(Y) = \set{y \in Y: \exists \epsilon > 0\text{ such that }B(y,\epsilon)\subseteq Y} \]
Define the boundary: \[\mathrm{Bd}(Y) = X \backslash (\mathrm{Int}(Y)\cup \mathrm{Int}(Y^c))\]
A subset \(Y\) in \((X,d)\) is open if \(Y=\mathrm{Int}(Y)\)
A subset \(Y\) in \((X,d)\) is closed if \(Y^c\) is open.
Let \((X,d)\) be a metric space, and let \(Y\subseteq X\). Then \(\mathrm{Int}(\mathrm{Int}(Y)) = \mathrm{Int}(Y)\).
For a subset \(Y\) of a metric space \((X,d)\), the set \(\mathrm{Int}(Y)\) is open.
The closure of \(Y\) is \(\mathrm{Cl}(Y) = \mathrm{Int}(Y) \sqcup \mathrm{Bd}(Y)\).
\(Y\) is dense if \(\mathrm{Cl}(Y) = X\)
Let \((X,d)\) be a metric space. An open neighbourhood of a point \(x\in X\) is an open set \(V\subseteq X\) such that \(x\in V\). A neighbourhood of \(x\) is a set \(U\subseteq X\) such that there is an open neighbourhood \(V\) of \(x\) with \(V\subseteq U\).
The set of open sets in a metric space \(X\) is called the topology of \(X\). \[\tau = \mathcal{O}(X)\]
Let \((X,d)\) be a metric space. The topology has the following properties:
- \(\varnothing, X \in \mathcal{O}(X)\)
- If \(\set{V_i}_{i\in I} \subseteq \mathcal{O}(X)\), then \(\displaystyle \bigcup_{i\in I} V_i \in \mathcal{O}(X)\). 𐃏
- If \(V_1, V_2 \in \mathcal{O}(X)\), then \(V_1 \cap V_2 \in \mathcal{O}(X)\) 𐃏
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. A function \(f:X\rightarrow Y\) is continuous if for every \(V\in \mathcal{O}(Y)\) we have \(f^{-1}(V)\in\mathcal{O}(X)\).
Let \((X,d_X)\), \((Y,d_Y)\) and \((Z,d_Z)\) be metric spaces. If \(f:X\to Y\) and \(g:Y\to Z\) are continuous, then the composition \[ g\circ f:X\longrightarrow Z,\qquad x\mapsto g\left (f(x)\right ), \] is continuous.
Let \((X,d)\) be a metric space and \(\varnothing\ne Y\subseteq X\). The following statements are equivalent:
- \(\forall x\in X\) there exists \(R(x)>0\) with \(Y\subseteq B(x,R(x))\);
- \(\exists\,y\in Y\) and \(R>0\) with \(Y\subseteq B(y,R)\);
- \(\exists\,R>0\) such that \(d(y_1,y_2)<R\) for every \(y_1,y_2\in Y\).
A subset \(Y\subseteq X\) satisfying the equivalent conditions above is bounded. (If \(Y=X\), we say the metric space itself is bounded.)
A sequence \(\{x_n\}_{n=0}^\infty\) in a metric space \((X,d)\) is Cauchy if \[ \forall\varepsilon>0\;\exists K(\varepsilon)\in\mathbb N\;:\; d(x_m,x_n)<\varepsilon\quad\text{whenever }m,n>K(\varepsilon). \]
A metric space \((X,d)\) is complete if every Cauchy sequence in \(X\) converges to a point of \(X\).
Let \((X,d)\) be complete and \(Y\subseteq X\) with the sub‑space metric. Then \(Y\) is complete iff \(Y\) is closed in \(X\).
The metric space \(\bigl(C[0,1],d_\infty\bigr)\) is complete.
Two Cauchy sequences \(\{a_n\}\) and \(\{b_n\}\) in \((X,d)\) are equivalent if \[ \lim_{n\to\infty} d(a_n,b_n)=0. \]
Let \(\overline X\) be the set of equivalence classes of Cauchy sequences in \(X\). For classes \([a_n]\) and \([b_n]\) define \[ \overline d\bigl([a_n],[b_n]\bigr)\;=\;\lim_{n\to\infty} d(a_n,b_n). \] Then \((\overline X,\overline d)\) is called the completion of \(X\).
Let \((X,d)\) be any metric space.
- \((\overline X,\overline d)\) is complete.
- The embedding \(i:X\to\overline X\), \(x\mapsto\) constant sequence \([x,x,\dots]\), is an isometry and \(i(X)\) is dense in \(\overline X\).
- The completion is unique: if \(j:X\to Y\) is an isometric embedding into a complete \(Y\) with dense image, then there is a unique bijective isometry \(f:Y\to\overline X\) satisfying \(f\circ j=i\).
For a vector space \(V\) (over \(\mathbb R\) or \(\mathbb C\)), a norm is a map \[ \|\cdot\|:V\to[0,\infty) \] such that for all \(x,y\in V\) and \(\lambda\in\mathbb R\text{ or }\mathbb C\):
- \(\|x\|=0\iff x=0\);
- \(\|\lambda x\|=|\lambda|\,\|x\|\);
- \(\|x+y\|\le\|x\|+\|y\|\).
If \((V,\|\cdot\|)\) is a normed space, then \[ d_{\|\cdot\|}(x,y)=\|x-y\|,\qquad x,y\in V, \] defines a metric on \(V\).
A Banach space is a normed vector space that is complete in the metric induced by its norm.
For every \(p\in[1,\infty)\) the space \[ \ell^p=\Bigl\{\,\{x_n\}_{n=1}^\infty\subset\mathbb R:\; \sum_{n=1}^{\infty}|x_n|^{\,p}<\infty\Bigr\}, \] with norm \(\|x\|_p=\bigl(\sum_{n}|x_n|^{\,p}\bigr)^{1/p}\), is a Banach space.
An inner product space is a vector space \(V\) together with \(\langle\cdot,\cdot\rangle:V\times V\to\mathbb R\text{ or }\mathbb C\) satisfying
- \(\langle x,x\rangle>0\) for \(x\ne0\);
- \(\langle x,y\rangle=\langle y,x\rangle\);
- \(\langle x+\lambda y,z\rangle=\langle x,z\rangle+\lambda\langle y,z\rangle\).
A Hilbert space is a complete inner product space.
A map \(f:(X,d)\to(X,d)\) is a contraction if \(\exists\,c\in(0,1)\) such that \(d \left (f(x),f(y)\right )\le c\,d(x,y)\) for all \(x,y\in X\).
If \(f\) is a contraction on \((X,d)\) and \(x_0\in X\), the sequence defined recursively by \(x_{n+1}=f(x_n)\) is Cauchy.
Let \((X,d)\) be complete and \(f:X\to X\) a contraction. Then \(f\) has a unique fixed point \(x^\ast=f(x^\ast)\), and for any \(x_0\in X\) the iteration \(x_{n+1}=f(x_n)\) converges to \(x^\ast\).
For \(X\subseteq\mathbb R\), a function \(f:X\to\mathbb R\) is Lipschitz continuous if \(\exists K>0\) such that \(|f(x)-f(y)|\le K|x-y|\) for all \(x,y\in X\). Such a constant \(K\) is called a Lipschitz constant for \(f\).
For \(X\subseteq\mathbb R^{\,2}\), a function \(f:X\to\mathbb R\) is Lipschitz continuous in the second variable if \(\exists K>0\) such that \[|f(x,y_1)-f(x,y_2)|\le K\,|y_1-y_2|\quad\forall (x,y_1),(x,y_2)\in X.\]
Let \(g\) be continuous on a neighbourhood of \((a,b)\in\mathbb R^{\,2}\) and Lipschitz in its second variable. Then there exists an interval about \(x=a\) on which the initial‑value problem \[ y’ = g(x,y),\qquad y(a)=b, \] has a unique solution.
Sequences and Series of Functions
A sequence of numbers \(\{x_n\}_{n=1}^{\infty}\subset\mathbb{R}\) converges to \(x\) if for every \(\varepsilon>0\) there exists \(K(\varepsilon)\in\mathbb{N}\) such that \[|x_n-x|<\varepsilon\quad\text{whenever }n\ge K(\varepsilon).\]
A sequence of functions \(f_n:X\to\mathbb{R}\) converges pointwise to \(f\) if for every \(x\in X\) and every \(\varepsilon>0\) there exists \(K(x,\varepsilon)\in\mathbb{N}\) such that \[|f_n(x)-f(x)|<\varepsilon\quad\text{whenever }n\ge K(x,\varepsilon).\]
A sequence of functions \(f_n:X\to\mathbb{R}\) converges uniformly to \(f\) if for every \(\varepsilon>0\) there exists \(K(\varepsilon)\in\mathbb{N}\) such that \[|f_n(x)-f(x)|<\varepsilon\quad\text{for all }x\in X\text{ whenever }n\ge K(\varepsilon).\]
Let \(B(X,\mathbb{R})\) be the set of bounded real‑valued functions on \(X\). The uniform norm is defined by \[\|f\|_\infty=\sup_{x\in X}|f(x)|.\]
\(\bigl(B(X,\mathbb{R}),\|\cdot\|_\infty\bigr)\) is a Banach space.
If \(E\) is a Banach space, then \(B(X,E)\) (bounded \(E\)‑valued functions on \(X\)) is Banach with the uniform norm, and if \(X\) is a metric space so is \(C_b(X,E)\).
Let \(f_n:[a,b]\to\mathbb{R}\) be Riemann‑integrable and \(p\ge1\). We say \(f_n\to f\) in \(L^p\) if \[ \lim_{n\to\infty}\int_a^b |f_n(x)-f(x)|^p\,dx=0. \]
Let \(E\) be a Banach space and \(\{x_n\}_{n=0}^{\infty}\subset E\). If the series of norms \(\sum_{n=0}^{\infty}\|x_n\|\) converges, then the series \(\sum_{n=0}^{\infty}x_n\) converges in \(E\).
Let \(f_n:X\to\mathbb{R}\) and let \(M_n\ge0\) satisfy \(|f_n(x)|\le M_n\) for all \(x\in X\). If \(\sum_{n=0}^{\infty}M_n\) converges, then the series \(\sum_{n=0}^{\infty}f_n(x)\) converges uniformly on \(X\).
Let \(f_n\in C[a,b]\) with \(f_n\to f\) uniformly. Then \[ \int_a^b f_n(x)\,dx\;\longrightarrow\;\int_a^b f(x)\,dx . \]
Let \(f_n\in C[a,b]\) be differentiable on \((a,b)\) with continuous, bounded derivatives \(f_n’\). If \(f_n\to f\) uniformly and \(f_n’\to g\) uniformly on \((a,b)\), then \(f\) is differentiable on \((a,b)\) and \(f’=g\).
If \(\sum_{n=0}^{\infty}f_n\) and \(\sum_{n=0}^{\infty}f_n’\) both converge uniformly on \((a,b)\), then the limit function is differentiable and \[ \left(\sum_{n=0}^{\infty}f_n\right)’=\sum_{n=0}^{\infty}f_n’. \]
For a power series \(\sum_{n=0}^{\infty}a_n x^n\) let \[ b=\limsup_{n\to\infty}|a_n|^{1/n},\qquad R=\frac1b\,(R\in[0,\infty]). \] The number \(R\) is the radius of convergence. (If \(b=0\) we set \(R=\infty\), while if \(b=\infty\) we set \(R=0\).)
With the notation above the power series converges absolutely when \(|x|<R\) and diverges when \(|x|>R\).
The term‑wise derivative \(\sum_{n=1}^{\infty}n\,a_n x^{n-1}\) has the same radius of convergence \(R\).
If \(\sum_{n=0}^{\infty}a_n x^n\) has radius of convergence \(R>0\), then it is differentiable on \((-R,R)\) and \[ \frac{d}{dx}\left(\sum_{n=0}^{\infty}a_n x^n\right)=\sum_{n=1}^{\infty}n\,a_n x^{n-1}. \]
Topological Spaces
A topological space is a pair \(\left( X ,\tau \right)\), where \(X\) is a set and \(\tau\subseteq\mathcal{P}\!\left( X \right)\) satisfies
- \(\varnothing , X \in \tau\);
- if \(\{V_i\}_{i\in I}\subseteq\tau\) then \(\bigcup_{i\in I} V_i \in \tau\);
- if \(V_1,V_2\in\tau\) then \(V_1\cap V_2\in\tau\).
The sets in \(\tau\) are open; their complements are closed.
For a topological space \(\left( X ,\tau \right)\) and \(Y\subseteq X\), the subspace topology on \(Y\) is \[ \tau\!\mid_Y=\{V\cap Y : V\in\tau\}. \]
In \(\left( X ,\tau \right)\) a subset \(C\subseteq X\) is closed if \(X\setminus C\in\tau\).
Let \(\left( X ,\tau \right)\) be a topological space.
- An open neighbourhood of \(x\in X\) is a set \(V\in\tau\) with \(x\in V\). A neighbourhood of \(x\) is any set containing an open neighbourhood of \(x\).
- For \(Y\subseteq X\) the interior is
\[ \operatorname{Int}\left( Y \right)=\{y\in Y : \exists V\in\tau,\; y\in V\subseteq Y\}. \]
For any \(Y\subseteq X\) the set \(\operatorname{Int}\left( Y \right)\) is open.
For \(Y\subseteq X\) set \[ \operatorname{Bd}\left( Y \right)=X\setminus\!\left( \operatorname{Int}\left( Y \right)\cup\operatorname{Int}\left( X\setminus Y \right) \right),\qquad \operatorname{Cl}\left( Y \right)=\operatorname{Int}\left( Y \right)\cup\operatorname{Bd}\left( Y \right). \]
A sequence \(\{x_n\}_{n=1}^{\infty}\subseteq X\) converges to \(x\in X\) if for every \(V\in\tau\) with \(x\in V\) there exists \(K(V)\in\mathbb{N}\) such that \(x_n\in V\) whenever \(n\ge K(V)\).
For topological spaces \(\left( X ,\tau_X \right)\) and \(\left( Y ,\tau_Y \right)\), a function \(f:X\to Y\) is continuous if \(f^{-1}\!\left( V \right)\in\tau_X\) for every \(V\in\tau_Y\).
If \(f:X\to Y\) and \(g:Y\to Z\) are continuous, then \(g\circ f:X\to Z\) is continuous.
A topological space is Hausdorff if for every distinct \(x,y\in X\) there exist disjoint neighbourhoods \(V\in\operatorname{Nbhd}\!\left( x \right)\) and \(U\in\operatorname{Nbhd}\!\left( y \right)\).
Let \(\left( X ,\tau \right)\) be a topological space.
- A base \( \mathcal{B}\subseteq\tau \) satisfies: every \(V\in\tau\) can be written \(V=\bigcup_{i\in I} B_i\) with \(B_i\in\mathcal{B}\).
- A local base at \(x\in X\) is a collection \(\mathcal{L}_x\subseteq\tau\) of neighbourhoods of \(x\) such that for every neighbourhood \(U\) of \(x\) there is \(V\in\mathcal{L}_x\) with \(V\subseteq U\).
Let \(X\) be a set and \(\mathcal{B}\subseteq\mathcal{P}\!\left( X \right)\). Define \(\tau=\{V\subseteq X : V\text{ is a union of sets in }\mathcal{B}\}\). Then \(\tau\) is a topology iff
- \(\bigcup_{B\in\mathcal{B}} B = X\);
- for every \(B_1,B_2\in\mathcal{B}\) and \(x\in B_1\cap B_2\) there exists \(B\in\mathcal{B}\) with \(x\in B\subseteq B_1\cap B_2\).
Given \(S\subseteq\mathcal{P}\!\left( X \right)\), let \(\mathcal{B}\) be the set of all finite intersections of elements of \(S\) (including \(X\)). The topology generated by \(S\) is \(\tau\!\left( S \right)=\{V\subseteq X : V\text{ is a union of sets in }\mathcal{B}\}\). The collection \(S\) is a subbase for \(\tau\!\left( S \right)\).
A space is first countable if every point has a countable local base. It is second countable if it possesses a countable base.
A topological space is separable if it contains a countable dense subset.
In a first countable space \(\left( X ,\tau \right)\) a subset \(Y\subseteq X\) is closed iff whenever a sequence in \(Y\) converges, its limit lies in \(Y\).
A directed set is a set \(\Lambda\) with a relation \(\le\) such that
- \(i\le i\) for all \(i\in\Lambda\);
- \(i\le j\le k\Rightarrow i\le k\);
- for \(i,j\in\Lambda\) there exists \(m\in\Lambda\) with \(i\le m\) and \(j\le m\).
A net in \(X\) is a function \(\Lambda\to X\) where \(\Lambda\) is directed; we write \(\{x_\lambda\}_{\lambda\in\Lambda}\).
A net \(\{x_\lambda\}\) converges to \(x\in X\) if for every neighbourhood \(V\) of \(x\) there exists \(\alpha\in\Lambda\) such that \(x_\lambda\in V\) whenever \(\lambda\ge\alpha\).
A subset \(Y\subseteq X\) is closed iff it contains the limits of all nets in \(Y\) that converge in \(X\).
For topologies \(\tau\subseteq\sigma\) on the same set \(X\), \(\tau\) is coarser and \(\sigma\) is finer.
A bijection \(f:X\to Y\) between topological spaces is a homeomorphism if both \(f\) and \(f^{-1}\) are continuous. Spaces that admit a homeomorphism are homeomorphic.
A space \(\left( X ,\tau \right)\) is connected if it cannot be written as the union of two disjoint non‑empty open sets. A subset \(Y\subseteq X\) is connected in the subspace topology.
If \(f:X\to Y\) is continuous and \(X\) is connected, then \(f\!\left( X \right)\) is connected.
Let \(\{W_i\}_{i\in I}\) be connected subsets of \(X\) with \(\bigcap_{i\in I} W_i\neq\varnothing\). Then \(\bigcup_{i\in I} W_i\) is connected.
A space is path‑connected if for all \(x,y\in X\) there exists a continuous map \(f:[0,1]\to X\) with \(f(0)=x\) and \(f(1)=y\).
Path‑connected \(\Rightarrow\) connected.
For topological spaces \(\left( X ,\tau_X \right)\) and \(\left( Y ,\tau_Y \right)\) the product topology on \(X\times Y\) is generated by the base \(\{U\times V : U\in\tau_X,\; V\in\tau_Y\}\).
Given \(\{(X_i,\tau_i)\}_{i\in I}\), the box topology on \(\prod_{i\in I} X_i\) has base \[ \left\{ \prod_{i\in I} U_i : U_i\in\tau_i\text{ for every }i\in I \right\}. \]
Compactness
A topological space \(\left( X ,\tau \right)\) is compact if for every open cover \[ X \;=\;\bigcup_{i\in I} V_i ,\qquad V_i\in\tau , \] there exists a finite sub‑cover \(V_{i_1},\ldots,V_{i_n}\) with \[ X \;=\;\bigcup_{k=1}^{n} V_{i_k}. \] A subset \(Y\subseteq X\) is compact when it is compact in the subspace topology.
The interval \([0,1]\) is compact.
If \(\left( X ,\tau_X \right)\) and \(\left( Y ,\tau_Y \right)\) are compact, then \(X\times Y\) is compact (with the product topology).
Rectangles of the form \([a,b]^n\subset\mathbb{R}^n\) are compact.
A subset \(X\subset\mathbb{R}^n\) is compact iff it is closed and bounded.
Every bounded sequence of real numbers has a convergent subsequence.
Every sequence of real numbers possesses a monotone subsequence.
Every bounded sequence in \(\mathbb{R}^n\) has a convergent subsequence.
A space \(\left( X ,\tau \right)\) is sequentially compact if every sequence in \(X\) has a convergent subsequence. The notion for subsets uses the subspace topology.
For \(X\subset\mathbb{R}^n\) the following are equivalent:
- \(X\) is compact;
- \(X\) is sequentially compact;
- \(X\) is closed and bounded.
If \(f:X\to Y\) is continuous and \(X\) is compact, then \(f\!\left( X \right)\) is compact.
A continuous function \(f:[a,b]\to\mathbb{R}\) attains its maximum and minimum.
For metric spaces \(\left( X ,d_X \right)\) and \(\left( Y ,d_Y \right)\) a function \(f:X\to Y\) is uniformly continuous if \[ \forall\varepsilon>0\;\exists\delta(\varepsilon)>0\text{ such that }d_Y\!\left( f(x),f(x’) \right)<\varepsilon \text{ whenever }d_X\!\left( x,x’ \right)<\delta(\varepsilon). \]
If \(\left( X ,d \right)\) is compact and \(f:X\to\mathbb{R}\) is continuous, then \(f\) is uniformly continuous.
A metric space \(\left( X ,d \right)\) is totally bounded if for every \(\varepsilon>0\) there exist points \(x_1,\ldots,x_n\in X\) such that \[ X \;=\;\bigcup_{k=1}^{n} B\!\left( x_k,\varepsilon \right). \]
For a metric space \(\left( X ,d \right)\) the following are equivalent:
- \(X\) is compact;
- \(X\) is sequentially compact;
- \(X\) is complete and totally bounded.
Let \(\left( X ,d_X \right)\), \(\left( Y ,d_Y \right)\) be metric spaces and \(S\subset C\!\left( X ,Y \right)\).
- Pointwise equicontinuous: \(\forall x\in X,\;\forall\varepsilon>0,\;\exists\delta(x,\varepsilon)>0\) such that \(d_Y\!\left( f(x),f(x’) \right)<\varepsilon\) for every \(f\in S\) whenever \(d_X\!\left( x,x’ \right)<\delta(x,\varepsilon)\).
- Uniformly equicontinuous: \(\forall\varepsilon>0,\;\exists\delta(\varepsilon)>0\) satisfying the same inequality for all \(x,x’\in X\) and all \(f\in S\).
If \(X\) is compact, a family \(S\subset C\!\left( X ,Y \right)\) is pointwise equicontinuous iff it is uniformly equicontinuous.
A bounded subset of \(\left( C[0,1],\|\cdot\|_\infty \right)\) is totally bounded iff it is equicontinuous.
A subset of \(\left( C[0,1],\|\cdot\|_\infty \right)\) is compact iff it is closed, bounded and equicontinuous.
Every uniformly bounded, equicontinuous sequence of functions on \([a,b]\) has a uniformly convergent subsequence.
For any continuous function \(f\) on \([a,b]\) and \(\varepsilon>0\) there exists a polynomial \(p\) with \[ \|f-p\|_\infty<\varepsilon . \]
A set \(A\subset F\!\left( X ,k \right)\) (functions from \(X\) to a field \(k\)) is an algebra if it is a vector space under pointwise operations and closed under pointwise multiplication. It is unital when it contains the constant function \(1\).
If \(X\) is a compact Hausdorff space, then \(C\!\left( X ,\mathbb{R} \right)\) separates points of \(X\).
Let \(X\) be a compact Hausdorff space and \(A\subset C\!\left( X ,\mathbb{R} \right)\) a unital sub‑algebra. Then \(A\) is dense in \(C\!\left( X ,\mathbb{R} \right)\) (with \(\|\cdot\|_\infty\)) iff \(A\) separates points.
Let \(X\) be compact Hausdorff and \(A\subset C\!\left( X ,\mathbb{C} \right)\) a unital ‑sub‑algebra (closed under complex conjugation). Then \(A\) is dense in \(C\!\left( X ,\mathbb{C} \right)\) *iff it separates points.
The product \(\prod_{i\in I} X_i\) of any family of compact spaces is compact in the product topology.
Let \(H\) be a Hilbert space. The closed unit ball \[ \{x\in H : \|x\|\le1\} \] is compact in the weak topology.
Backlinks (3)
1. My Undergraduate Computer Science Degree /blog/ugrad-unsw/
Summary
It took me 1,577 days to complete my undergraduate degree in Computer Science.
Overall, I think the degree was worthwhile and satiated my desire to obtain an understanding of Computers from the bit-level.
2. Wiki /wiki/
Knowledge is a paradox. The more one understand, the more one realises the vastness of his ignorance.

