lacunary - Theorem Index

Since $\mathbb{C}$ is a field, ${\mathbb{C}}^N$ is obviously a vector space.

To see that this is indeed an inner product, we'll show the required properties hold.

For @conjugate-symmetry, let $z_k=a+bi$ and let $w_k=c+di.$ Then,

$$z_k\overline{w_k}=(a+bi)(c-di)=ac-adi+cbi+bd=(ac+bd)+(cb-ad)i,$$

and

$$\overline{w_k\stackrel{―}{z_k}}=\overline{(c+di)(a-bi)}=\overline{ac+adi-cbi+bd}=(ac+bd)-(ad-cb)i=(ac+bd)+(cb-ad)i,$$

and since $k$ is arbitrary, it holds for any respective pair of components in $z$ and $w,$ and so $⟨z,w⟩=\overline{⟨w,z⟩}.$

For @linearity in the first argument, let $w,v,z\in {\mathbb{C}}^n,a,b\in \mathbb{C}.$ Then

$$⟨ax+by,z⟩=(a{x}_1+b{y}_1)\ast z_1+(a{x}_2+b{y}_2)\ast z_2+\cdots =a(x_1\ast z_1)+a(x_2\ast z_2)+\cdots +b(y_1\ast z_1)+b(y_2\ast z_2)+\cdots =a⟨x,z⟩+b⟨x,z⟩.$$

Now, to show @positive-definiteness let $z=a+bi.$ Then,

For (a), assume ${\mathbf{x}}_n\to \mathbf{x}.$ Then, from the definition of the norm,

$$|{\alpha }_{j,n}-{\alpha }_j|\le |{\mathbf{x}}_n-\mathbf{x}|,$$

that is, the distance from ${\alpha }_{k,n}$ to ${\alpha }_n$ is always less than or equal to the distance from ${\mathbf{x}}_n$ to $\mathbf{x}.$ Therefore, for and we can pick $n$ to make this true for as small of $ϵ$ as we'd like. Therefore, $\underset{n\to \mathrm{\infty }}{lim}{\alpha }_{j,n}={\alpha }_j.$

Conversely, assume $\underset{n\to \mathrm{\infty }}{lim}{\alpha }_{j,n}={\alpha }_j.$ Let For some integer $N,$ when $n\ge N$ we have

$$|{\alpha }_{j,n}-{\alpha }_j|\le \frac{ϵ}{\sqrt{k}},(1\le j\le k).$$

Therefore, $n\ge N$ implies that

so ${\mathbf{x}}_n\to \mathbf{x}.$

Part (b) follows from part (a) and A sequence in $R^k$ converges iff its components converge.

By assumption and definition of the complex derivative, we have that

Now, this limit must exist and be the same no matter what path we take. So, we can take the path along the real axis by setting $\mathrm{\Delta }y=0$ and sending $\mathrm{\Delta }x\to 0,$ and we get

$$f^{\prime }(z_0)=f^{\prime }(x_0+i{y}_0)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{[u(x_0+\mathrm{\Delta }x,y_0)-u(x_0,y_0)]+i[v(x_0+\mathrm{\Delta }x,y_0)-v(x_0,y_0)]}{\mathrm{\Delta }x},$$

and hence,

$$\mathrm{Re}{f}^{\prime }(z_0)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{u(x_0+\mathrm{\Delta }x,y_0)-u(x_0,y_0)}{\mathrm{\Delta }x},$$

$$\mathrm{Im}{f}^{\prime }(z_0)=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{v(x_0+\mathrm{\Delta }x,y_0)-v(x_0,y_0)}{\mathrm{\Delta }x}.$$

This gives us that the partial derivatives $u_x$ and $v_x$ exist and

$$\frac{\partial u}{\partial x}(x_0,y_0)=\mathrm{Re}{f}^{\prime }(x_0+i{y}_0),\frac{\partial v}{\partial x}(x_0,y_0)=\mathrm{Im}{f}^{\prime }(x_0+i{y}_0),$$

that is,

$$\begin{array}{cc}{f}^{\prime }(x_0+i{y}_0)=u_x+i{v}_x.& \text{(a)}\end{array}$$

Now, we can repeat this trick by sending $\mathrm{\Delta }z=\mathrm{\Delta }x+i\mathrm{\Delta }y$ to zero along the imaginary axis by setting $\mathrm{\Delta }x=0$ and sending $\mathrm{\Delta }y\to 0$ to get

and hence

$$\mathrm{Re}{f}^{\prime }(z_0)=\underset{\mathrm{\Delta }y\to 0}{lim}\frac{v(x_0,y_0+\mathrm{\Delta }y)-v(x_0,y_0)}{\mathrm{\Delta }y},$$

$$\mathrm{Im}{f}^{\prime }(z_0)=-\underset{\mathrm{\Delta }y\to 0}{lim}\frac{u(x_0,y_0+\mathrm{\Delta }y)-u(x_0,y_0)}{\mathrm{\Delta }y}.$$

This gives us that the partial derivatives $u_y$ and $v_y$ exist and

$$-\frac{\partial u}{\partial y}(x_0,y_0)=\mathrm{Im}{f}^{\prime }(x_0+i{y}_0),\frac{\partial v}{\partial y}(x_0,y_0)=\mathrm{Re}{f}^{\prime }(x_0+i{y}_0),$$

that is,

$$\begin{array}{cc}{f}^{\prime }(x_0+i{y}_0)=v_y-i{u}_y.& \text{(b)}\end{array}$$

Now, comparing (a) and (b) we have

$$u_x+i{v}_x=v_y-i{u}_y,u_x=v_y,u_y=-v_x.$$

Let $\mathbf{y},\mathbf{z}$ be points in a ball with center $\mathbf{x}$ and radius $r$. Then, by definition, and Suppose We will show that

So, $\mathbf{p}$ is within our ball and therefore all balls are convex.

Let $E$ be a countable subset of $A,$ and call the elements of $E$ $s_1,s_2,s_3,\dots .$ We will construct a new sequence $p$ in the following way:

$$p_n=\mathrm{\neg }{s}_{n_n},n=1,2,3,\dots$$

That is, the $n$th digit of $p$ will be the opposite of whatever the $n$th digit of $s_n$ is. So, $p$ differs from $s_1$ in the first digit, from $s_2$ in the second digit, $s_3$ in the third digit, and so on, so that it differs from all elements of $E,$ and therefore is not contained in $E.$ But, $p$ is definitely in $A$ since it its elements are the digits $0$ and $1.$ Therefore, $E$ is a proper subset of $A,$ so any countable subset of $A$ must be a proper subset of $A.$ But, $A$ can't be a proper subset of itself, and therefore $A$ must be uncountable.

Note that any bounded sequence $\{p_n\}\subset R^k$ is a subset of some closed set, bounded and thus compact k-cell in $R^k.$ Therefore, $\{p_n\}$ is a sequence in a compact metric space, and has a convergent subsequence.

Let $X$ be a separable metric space and $E\subset X$ be closed. If $E$ is at most countable, then we are done.

Suppose that $E$ is uncountable. Note that the proof of Suppose $E\subset R^k,$ with $E$ uncountable,... only uses the property that $R^k$ is a separable metric space, and it therefore generalizes to any separable metric space. Thus $E$ is the union of a perfect set - its condensation points, $P$, and a set that is at most countable, $E\setminus P.$

Suppose, for the sake of contradiction, that some segment $(\alpha ,\beta )\subset P$ and let $L=\beta -\alpha .$ Pick some $n\in \mathbb{N}$ such that Now, $E_n$ is the union of $2^n$ intervals of length $1/3^n,$ and since $(\alpha ,\beta )\subset P,$ it must be the case that $(\alpha ,\beta )$ is a subset of some interval of length $1/3^n.$ However, this can't be the case, since by construction. Therefore, our provision assumption is incorrect, and $P$ contains no segment.

Clearly, $P$ is bounded, for it lies within $[0,1].$ Each $E_n$ is composed of the union of $2^n$ closed intervals, and the union of finitely many closed intervals is also closed. $P$ is then the intersection of infinitely many closed intervals, which is again closed. Therefore, $P$ is closed and bounded, and by Heine-Borel, is compact.

Suppose $E_n$ has $[\alpha ,\beta ]$ as an interval and thus $\alpha ,\beta \in E_n.$ Then, by definition, $E_{n+1}$ will contain $[\alpha ,\frac{\beta -\alpha }{3}]$ and $\frac{2(\beta -\alpha )}{3},\beta ]$ as intervals, so $\alpha ,\beta \in E_{n+1}.$ Note that $E_0$ has $[0,1]$ as an interval. By induction, all $E_n$ contain $0$ and $1$, and therefore so does their intersection $P,$ and $P$ is nonempty.

Let $x\in P.$ Let and pick $n\in \mathbb{N}$ to be large enough that Then, $x$ lies in one of the $2^n$ intervals of length $1/3^n$ in $E_n;$ call it $I_n.$ The endpoints of $I_n$ are also in $P$ (see The Cantor set is not empty.,) and at least one of them is not $x.$ Since the endpoints are contained in a neighborhood of $x$ with radius all neighborhoods of $x$ are limit points of $P,$ and therefore $P$ is perfect.

Suppose $\{p_n\}\subset R^k$ converges. Then, because $R^k$ is a metric space, $\{p_n\}$ is cauchy.

Conversely, suppose $\{p_n\}\subset R^k$ is cauchy. Then, $\{p_n\}$ converges.

Let $f(z)=u(x,y)+i(v,y).$ By the definition of the contour integral, we have

$${\int }_{C}f(z)dz={\int }_C(u+vi)(dx+dyi)={\int }_C(udx-vdy)+i{\int }_C(vdx+udy).$$

Because $f$ is analytic, the Cauchy-Riemann Equations tell us that $u(x,y)$ and $v(x,y)$ have continuous first partial derivatives, and therefore, we can apply Green's Theorem. Now, we can take the first integral and transform it

$${\int }_C(udx+(-vdy))=-{\iint }_R(v_x+u_y)dxdy.$$

But, by Cauchy-Riemann Equations, $v_x=-u_y,$ so this integral evaluates to $0.$

Now, for the second integral, using Green's Theorem again, we get

$$i{\int }_C(vdx+udy)=i{\iint }_R(u_x-v_y)dxdy,$$

and by Cauchy-Riemann Equations, $u_x=v_y,$ so this integral also evaluates to $0,$ which makes the overall integral $0$ as well.

Note that

The first term on the RHS is just $-\sum _{x,y}p(x,y)\log p(x,y)=H(X,Y).$ The second is

Altogether, we have that $H(Y|X)=H(X,Y)-H(X),$ or

$$H(X,Y)=H(X)+H(Y|X).$$

Let $X$ be a random variable with mean $E[X]$ and variance ${\sigma }^2$. Define the non-negative random variable

$$Y=(X-E[X]{)}^2.$$

By Markov's Inequality, for any , we have

$$P(Y\ge a)\le \frac{E[Y]}{a}.$$

Set $a=t^2$. Then,

$$P((X-E[X]{)}^2\ge t^2)\le \frac{E[(X-E[X]{)}^2]}{t^2}.$$

Since the variance of $X$ is ${\sigma }^2=E[(X-E[X]{)}^2]$, it follows that

$$P((X-E[X]{)}^2\ge t^2)\le \frac{{\sigma }^2}{t^2}.$$

Note that

$$(X-E[X]{)}^2\ge t^2⟺|X-E[X]|\ge t.$$

Thus,

$$P(|X-E[X]|\ge t)\le \frac{{\sigma }^2}{t^2}.◻$$

Suppose $F\subset K\subset X,$ with $F$ closed relative to $X,$ and $K$ compact. Let $\{V_{\alpha }\}$ be an open cover of $F.$ Since $F^c$ is open relative to $X$ (see A set $E$ is open iff its...), if we add it to $\{V_{\alpha }\},$ we obtain an open cover of $K;$ let's call it $\mathrm{\Omega }.$ Since $K$ is compact, we can obtain a finite subcover of $K$ by discarding all but a finite number of sets from $\mathrm{\Omega };$ let's call it $\mathrm{\Phi }.$ Since $F\subset K,$ $\mathrm{\Phi }$ is also a finite subcover of $F,$ and therefore $F$ is compact.

Let $p\in \bigcup _{i=1}^n\overline{A_i}.$ Then, $p\in \overline{A_i}$ for some $i.$ If $p\in A_i,$ then $p\in \bigcup _{i=i}^n{A}_i=B_n,$ so $p\in \overline{B_n}.$ If $p$ is in some $A_i^{\prime },$ then every neighborhood of $p$ contains some point $q\in A_i,q\ne p,$ and since $A_i\subset B_n,$ $q\in B_n,$ so $p\in \overline{B_n}.$

Conversely, let $p\in \overline{B_n}.$ If $p\in B_n,$ $p\in A_i$ for some $i,$ and thus $p\in \overline{A_i}\subset \bigcup _{i=1}^n\overline{A_i}.$ If $p$ is only in $B_n^{\prime },$ suppose for contradiction that $p\notin \bigcup _{i=1}^n\overline{A_i}.$ For each $i,$ let $N_i$ be a neighborhood centered at $p$ with $N_i\cap A_i=\mathrm{\varnothing }.$ Then, $N=\bigcap _{i=1}^n{N}_i$ is a neighborhood of $p$ (see (a) - For any collection $\{G_a\}$ of...,) but $N\cap N_i=\mathrm{\varnothing }$ for all $i,$ so $N\cap B_n=\mathrm{\varnothing }.$ But, this contradicts our hypothesis that $p\in \overline{B_n},$ so our contradictory assumption must be invalid, and $p\in \bigcup _{i=1}^n\overline{A_i}.$

Suppose $K$ is compact relative to metric space $X.$ Let $p\in K^c.$ For each $q\in K,$ we can define $r_q=\frac{1}{2}d(p,q),$ and let $P_q$ and $Q_q$ be neighborhoods of radius $r_q$ around $p$ and $q,$ respectively. Note that $P_q$ and $Q_q$ are disjoint, because we defined their radii to be half the distance between them, and they are open.) Now, since $K$ is compact, we can pick a finite number of points in $K,$ $q_1,\dots ,q_n,$ such that $K\subset Q_{q_1}\cup \cdots \cup Q_{q_n}=Q$ ($Q$ is a finite subcover of $K.$) Using the same set of points as reference, let $P_{q_1}\cap \cdots \cap P_{q_n}=P.$ Note that $P\cap Q=\{\},$ since each $P_q$ is disjoint with its paired $Q_q$ (to be in $P$, a point must be in all $P_q,$ but any point in $Q$ is not in at least one $P_q.$) $P$ is open, since it is the intersection of finitely many open sets (see (a) - For any collection $\{G_a\}$ of...) and obviously contains $p,$ since each $P_q$ contains $p$. Therefore, $p$ has a neigborhood $P$ that is disjoint with $K$ (since $P\cap K\subset P\cap Q=\{\}$), and is therefore an interior point of $K^c.$ It follows that $K^c$ is open, and that $K$ is closed.

Let $K$ be a compact metric space. Fix $n\in \mathbb{N}$ (any natural number will do,) then, consider the open cover $\{G_{\alpha }{\}}_n$ of balls of radius $1/n$ centered at $x\in K.$ Now, because $K$ is compact, some finite $\{V_{\alpha }{\}}_n\subset \{G_{\alpha }{\}}_n$ covers $K.$

Now, consider the set of all balls for all $n,$

$$C=\bigcup _{n=1}^{\mathrm{\infty }}\{V_{\alpha }{\}}_n$$

Because the union of a sequence of countable sets is countable, $C$ is countable. To show $C$ is a base for $K,$ let $G$ be an open subset of $K,$ and let $x\in G.$ Let such that $B_{ϵ}(x)\subset G.$ Pick natural $n$ such that Then, $x\in \{V_{\alpha }{\}}_n$ for some $\alpha ,$ (because no point in $K$ is more than $1/n$ away from the center of some $\{V_{\alpha }{\}}_n$ and therefore $C$ is a countable base of $K.$ Since we can make $ϵ$ as close to $0$ as we like, the centers of $\{V_{\alpha }{\}}_n$ are dense as well as countable, and are therefore a countable dense subset of $K,$ so $K$ is separable.

Let $\{p_n\}$ be a cauchy sequence in the compact metric space $X.$ For $N=1,2,3,\dots ,$ let $E_n$ bet the set consisting of $p_N,p_{N+1},p_{N+2},\dots .$ Then

$$\underset{N\to \mathrm{\infty }}{lim}\mathrm{diam}\overline{E_N}=0,$$

by two theorems above. Each $\overline{E_n}$ is a closed subset of the compact space $X,$ and is thus compact. Also, $E_{N+1}\subset E_N,$ which implies that $\overline{E_{N+1}}\subset \overline{E_N}$. Now, we have that there is a unique $p\in X$ which lies in every $\overline{E_n}.$

Let Since $\underset{N\to \mathrm{\infty }}{lim}\mathrm{diam}\overline{E_N}=0,$ there is an integer $N_0$ such that if $N\ge N_0.$ Since $p\in \overline{E_n},$ we have that for every $q\in \overline{E_N}$ and thus for every $q\in E_n.$ That is, if $n\ge N_0,$ so $\{p_n\}$ converges to $p.$

Suppose $K$ is compact relative to $X$ and that $\{V_{\alpha }\}$ is an open cover of $K$ relative to $Y,$ such that $K\subset \bigcup _{\alpha }{V}_{\alpha }.$ We need to show that a finite subset of $\{V_{\alpha }\}$ covers $K.$ Because $\{V_{\alpha }\}$ is open relative to $Y$ and $Y\subset X$ (see Suppose $Y\subset X.$ A subset $E$...) there are sets $G_{\alpha },$ open relative to $X,$ such that $V_a=Y\cap G_{\alpha },$ for each $\alpha .$ Now, since $K$ is compact relative to $X,$ we have $$\begin{array}{cc}K\subset G_{{\alpha }_1}\cup \cdots \cup G_{{\alpha }_n},& \text{(a)}\end{array}$$ for some finite set of indices $a_1,\dots ,a_n.$ Now, since $K\subset Y,$ $$\begin{array}{cc}K\subset V_{{\alpha }_1}\cup \cdots \cup V_{{\alpha }_n},& \text{(b)}\end{array}$$ which shows $K$ is compact relative to $Y.$

Conversely, suppose $K$ is compact relative to $Y$ and let $\{G_{\alpha }\}$ be a cover of $K$ open relative to $X.$ We need to show there is a finite subset of $\{G_{\alpha }\}$ that covers $K.$ Let $V_{\alpha }=Y\cap G_{\alpha },$ for each $\alpha .$ Then (b) will hold for some set of indicies, ${\alpha }_1,\dots ,{\alpha }_n,$ and since each $V_{\alpha }\subset G_{\alpha },$ (a) is implied by (b) and we've shown $K$ is compact relative to $X.$

Let $A$ and $B$ be sets. We want to show that

$$(A\cap B{)}^c=A^c\cup B^c.$$

Suppose $x\in (A\cap B{)}^c.$ Then, $x$ is not in $A\cap B,$ that is, $x$ is either not in $A$ or it is not in $B$ or it is in neither. If $x$ is not in $A,$ then it is in $A^c,$ and therefore it is in $A^c\cup B^c.$ The same approach works with $B,$ and therefore $x\in A^c\cup B^c,$ and we have shown $(A\cap B{)}^c=A^c\cup B^c.$

Let $A$ and $B$ be sets. We want to show that

$$(A\cup B{)}^c=A^c\cap B^c.$$

Suppose $x\in (A\cup B{)}^c.$ Then, if $x\in A$ or $x\in B,$ then $x\in A\cup B$ and $x\notin (A\cup B{)}^c,$ a contradiction. Therefore, $x\notin A$ and $x\notin B.$ That is, $x\in A^c$ and $x\in B^c,$ therefore $x\in A^c\cap B^c.$

Suppose $x\in {\left(\bigcup _{\alpha }{E}_{\alpha }\right)}^c.$ Then, $x\notin \bigcup _{\alpha }{E}_{\alpha },$ so $x$ is not in any $E_{\alpha }.$ Therefore, for every $E_{\alpha },$ $x\in E_{\alpha }^c,$ and thus $x\in \bigcap _{\alpha }\left(E_{\alpha }^c\right).$ Conversely, suppose $x\in \bigcap _{\alpha }\left(E_{\alpha }^c\right).$ Then, $x$ is in every $E_{\alpha }^c,$ that is, $x$ is not in any $E_{\alpha }.$ Therefore, $x\in {\left(\bigcup _{\alpha }{E}_{\alpha }\right)}^c.$

Let Since $g$ is continuous at $f(p),$ there exists such that

Since $f$ is continuous at $p,$ there exists such that

It follows that

if and $x\in E.$ Thus, $h$ is continuous at $p.$

Let $\{V_n\}$ be a countable base of $R^k$ (see $R^k$ is separable. and Every separable metric space has a countable...,) and let $W$ be the union of those $V_n$ for which $E\cap V_n$ is at most countable. We will show that $P=W^c.$

Suppose $p\in W^c.$ Then $p$ is in no $V_n$ for which $V_n\cap E$ is at most countable, that is, every neighborhood of $p$ has uncountably many points in $E,$ and thus $p\in P.$

Conversely, suppose $p\in P.$ Suppose, for the sake of contradiction, that $p\in W.$ Then $p\in V_n$ for some $V_n$ where $V_n\cap E$ is at most countable. But, since $p$ is an interior point of this $V_n,$ there is a neighborhood $N(p)\subset V_n,$ and since every neighborhood of $p$ has uncountably many points in $E,$ we have a contradiction, and thus our assumption that $p\in W$ must be incorrect, and therefore $p\in W^c,$ and $P=W^c.$ Furthermore, since $W$ is a union of open sets, $W$ is open, and $W^c=P$ is closed.

Since $W$ is open, only countably many $V_n$ are required to cover it. Each of these $V_n$ has at most countably many points in $E,$ so $W=P^c$ has at most countably many points in $E,$ that is, there are at most countably many points of $E$ that are not in $P.$

Now, to show all points in $P$ are limit points of $P,$ suppose $p\in P.$ Let be a neighborhood of $p.$ Then, $N_r(p)\cap E$ is uncountable. Now, since there are at must countably many points in $E$ that are not in $P,$ there are at most countably many points in $(N_r(p)\cap E)\setminus P,$ and therefore there must be uncountably many points in $N_r(p)\cap E\cap P.$ Therefore, every neighborhood of $p$ contains infinitely many points in $P$ other than $p,$ $p$ is a limit point of $P,$ and $P$ is perfect.

We will proceed both sides of the implication by proving the contrapositive, i.e., that if the interval property doesn't hold, then the set isn't connected, and conversely, that if the set isn't connected, the interval property doesn't hold.

Suppose $x,y\in E$ and $z\in (x,y),z\notin E.$ Then $E=A_z\cup B_z,$ where

$$A_z=E\cap (-\mathrm{\infty },z),B_z=E\cap (z,\mathrm{\infty }).$$

Since $x\in A_z$ and $y\in B_z,$ they are nonempty, and since $A_z\subset (-\mathrm{\infty },z)$ and $B_z\subset (z,\mathrm{\infty }),$ they are separated. Therefore, $E$ is not connected.

Conversely, suppose, for the sake of contradiction, that $E$ is not connected. Then there are nonempty separated sets $A$ an $B$ such that $A\cup B=E.$ Let $x\in A,y\in B$ and assume Define

$$z=sup(A\cap [x,y]).$$

By Let $E$ be a nonempty set of..., $z\in \bar{A},$ and because $A$ and $B$ are separated, $z\notin B.$ Therefore

If $z\notin A,$ it follows that and $z\notin E.$

If $z\in A,$ then $z\notin \bar{B},$ hence there exists $z_1$ such that and $z_1\notin B$ (because $z\notin \bar{B}$ means there is a neighborhood of $z$ that contains no points of $B$.) Thus, and $z_1\notin E.$

Suppose ${\mathbf{x}}^{\mathbf{\ast }}$ were an attracting fixed point in a conservative system. Then, note that all trajectories in the basin of attraction approach ${\mathbf{x}}^{\mathbf{\ast }}$ as $t\to \mathrm{\infty }.$ Since $E(\mathbf{x})$ is continuous, the energy in the limit as trajectories approach ${\mathbf{x}}^{\mathbf{\ast }}$ is equal to the energy at the fixed point, and so the energy along the entirety of each trajectory is equal to the energy at the fixed point. But, this implies that the energy in the entire basin of attraction is the same as the energy at the fixed point, which violates our definition of a conservative system (i.e. we require that $E(\mathbf{x})$ be nonconstant.) Therefore no such fixed point can exist.