lacunary - Mathnotes

We know that $P$ is infinite, because by definition, all points in perfect sets are limit points, and only infinite sets have limit points.

Suppose, for the sake of contradiction, that $P$ is countable. Label the points of $P$ as $x_1,x_2,\dots .$ We will construct a sequence of $V_n$ of neighborhoods.

As a base step, let $V_1$ be any neighborhood of $x_1;$ let (note: subsequent $V_{n+1}$ aren't required to be neighborhoods of $x_{n+1}.$) Then the closure $\overline{V_1}$ of $V_1$ is $\overline{V_1}=\{y\in R^k||y-x_1|\le r\}.$

For the inductive step, suppose as an induction hypothesis that we have some $V_n$ that's been constructed such that $V_n\cap P$ is not empty. Since every point of $P$ is a limit point of $P,$ we can make a neighborhood $V_{n+1}$ such that (i) $\overline{V_{n+1}}\subset V_n,$ (ii) $x_n\notin \overline{V_{n+1}},$ (iii) $V_{n+1}\cap P$ is not empty. Now, $V_{n+1}$ satisfies our induction hypothesis, and since $V_1$ does too, we have $V_n$ defined for all $n=1,2,3,\dots .$

For each $n$, let $K_n=\overline{V_n}\cap P.$ Since $\overline{V_n}$ is closed and bounded, $\overline{V_n}$ is compact. Since $x_n\notin \overline{V_{n+1}},$ no point of $P$ lies in $\bigcap _{n=1}^{\mathrm{\infty }}{K}_n.$ Since $K_n\subset P,$ this implies that $\bigcap _{n=1}^{\mathrm{\infty }}{K}_n$ is empty. But, each $K_n$ is nonempty, by (iii), and $K_{n+1}\subset K$, by (i). But the intersection of nonempty compact nested sets is nonempty, so we have a contradiction, so our provisional assumption that $P$ is countable must be incorrect. Therefore, $P$ is uncountable.

The points of $R^k$ that have only rational coordinates are a subset of $R^k,$ we'll call it $Q^k,$ countable and dense.

We know that the rationals are countable, and because $n$-tuples of countable elements are countable, $Q^k$ is countable.

To show $Q^k$ is dense, consider an arbitrary point $p$ in $R^k.$ Now, let Since the the rationals are dense in the reals, we can pick a $q\in Q^k$ with by picking rational approximations of the coordinates of $p$ and forming $q$ such that $q$ is within $ϵ$ of $p,$ i.e.

Let $X$ be a separable metric space, and let $x\in G\subset X,$ with $G$ @open-set. Now, because $X$ is separable, it by definition has a countable dense subset $E.$ If $x\in E,$ we can pick a rational and let $N_{\delta }(x)$ be a neighborhood such that $N_{\delta }(x)\subset G$ ($\delta$ must also be small enough such that this neighborhood is within $G,$ which is possible because $x$ is an interior point of $G$ and we can pick a rational as close to any real as we'd like.) Now we have that $x\in N_{\delta }(x)\subset G,$ and since there are countably many $x\in E$ and countably many neighborhoods with rational radius around each $x\in E,$ there are countably many such neighborhoods in $G.$ On the other hand, if $x\notin E,$ then $x$ is a limit point of $E,$ and thus there is $p\in E$ as close as we'd like to $x.$ Pick $p\in E$ such that for some rational $\delta$ such that $x\subset N_{\delta }(p)\subset G.$ Again, since there are countably many such $p\in E,$ with countably many neighborhoods of rational radius each, there are countably many such neighborhoods in $G.$

Now, if we let $G$ be the union of all open sets $G_{\beta }\subset X,$ then $G$ is open, and let $V_{\alpha }$ be the union of all the $N_r(q),q\in G\cap E,$ with $r\in Q,$ then every $x\in G$ is in some $V_{\alpha },$ and there are countably many $V_{\alpha },$ so $V_{\alpha }$ is a countable base for $X.$

Let and pick $x_1\in X.$ Now, continue picking $x_{j+1}$ such that $d(x_i,x_{j+1})\ge \delta$ i.e. so that each new point is at least $\delta$ away from each existing point. Suppose this process continues infinitely; then the points $\{x_i\}$ are an infinite subset of $X,$ and thus must have a limit point $p.$ Now, because every neighborhood of $p$ must contain infinitely many points in $x_i$, we can let $r=\delta /2,$ $N_{r}p$ be a neighborhood of $p,$ and pick $x,y\in \{x_i\},N_{r}p.$ But, by the triangle inequality

so which contradicts our assumption that $\{x_i\}$ could be an infinite set where all points were at least $\delta$ apart. Therefore, $\{x_i\}$ has only finitely many points, and $X$ can be covered with finitely many open balls of radius $\delta$ (for if it couldn't be, we could always fit another point into $\{x_i\}.$)

Now, if we let $\delta =1/n,n=1,2,3,\dots ,$ we can consider the set of points $\{x_i{\}}_n$ as the finite set of points at least $1/n$ apart in $X.$ The collection of all such points can be called $\{x_{i_n}\}$ and is dense in $X:$ let $p$ be a point in $X.$ and a neighborhood of $p.$ If we pick $n$ such that then some $\{x_i{\}}_n$ will be in $N_{r}p,$ because if not, we would have a contradiction with the fact that shown above that no point in $X$ is more than $1/n$ away from a point in $\{x_i{\}}_n.$

Since each $\{x_i{\}}_n$ is finite, and there are countably many $\{x_i{\}}_n,$ the $\{x_{i_n}\}$ is a countably dense subset of $X$ and therefore $X$ is separable.

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.

By Let $X$ be a metric space in..., we know that $X$ is separable and by Every separable metric space has a countable..., we know that $X$ has a countable base. By the definition of base, we have that every open cover of $X$ has a countable subcover $\{G_n\},n=1,2,3,\dots .$

Suppose, for the sake of contradiction, that $\{G_n\}$ has no finite subcollection that covers $X.$ Let

$$F_n=(G_1\cup \cdots \cup G_n{)}^c.$$

Then, each $F_n$ must be nonempty (otherwise a finite subcollection of $\{G_n\}$ would cover $X.$ However, since every point in $X$ is in some $G_n,$

$$\bigcap _{n=1}^{\mathrm{\infty }}{F}_n=\mathrm{\varnothing }.$$

Now, let $E$ be a set with a point from each $F_n.$ Since there are infinitely many $F_n,$ $E$ is an infinite subset of $X,$ and therefore $E$ has a limit point, $p.$ Now, $p$ must be in some open $G_m,$ and so for some $N_{ϵ}(p)\subset G_m.$

Now, note that each $F_{n+1}\subset F_n,$ because $F_{n+1}$ is formed by excluding all the points in $F_n$ that are also in $G_{n+1}.$ Because $p$ is a limit point of $E,$ $N_{ϵ}p$ must contain infinitely many points of $E.$ However, only finitely many points of $E$ can be in $N_{ϵ}(p)\subset G_m,$ because for $F_n$ contains no points in $G_m.$ Therefore, $p$ is not a limit point of $E,$ and no such limit point can exist, contradicting our hypothesis that every infinite subset of $X$ has a limit point. Therefore, $E$ must be finite, and $\{G_n\}$ must have a finite subcollection that covers $X,$ meaning $X$ is compact.

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.

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, $E\setminus P.$

Let $E$ be an open set in $R^1.$ Let $x\in E.$ Let $a=inf\{y:(y,x]\subset E\}$ and $b=sup\{y:[x,y)\subset E\}.$ Then, because $E$ is open and thus $x$ is an interior point of $E$ and there is some neighborhood around $x$ that is entirely within $E.$ Let $I(x)=(a,b).$ By construction, $I(x)$ is connected. For all $u\in I(x),$ $I(u)$ must have the same end points as $I(x),$ since if it extended beyond, our construction of $I(x)$ would be contradicted. Also note that for all $v\in E\setminus I(x),$ $I(v)\cap I(x)=\mathrm{\varnothing },$ for if they intersected, they would form an open interval. Thus, each $I(x),x\in E$ is either disjoint from all others or identical to some other $I(y),y\in E,$ and $E$ is the union of all such unique $I.$

Now, in each $I,$ we can pick a rational number. Because the rationals are countable, we have at most countably many unique $I,$ and $E$ is their union.

First, note that since $R^k=\bigcup _{n=1}^{\mathrm{\infty }}{F}_n,$ every point in $R^k$ is in some $F_n,$ and so $\bigcap _{n=1}^{\mathrm{\infty }}{F}_n^c$ must be empty.

Suppose, for the sake of contradiction, that no $F_n$ has a non-empty interior, that is, every $F_n$ is closed with an empty interior. Then, each $F_n^c$ is open. Moreover, since $F_n$ has an empty interior, it has no points for which there exists a neighborhood that contains only points in $F_n,$ that is, every neighborhood of each point of $F_n$ contains a point in $F_n^c,$ so every point in $F_n$ is a limit point of $F_n^c,$ and every point in $R^k$ is either in $F_n^c$ or is a limit point of $F_n^c,$ so each $F_n^c$ is non-empty and dense in $R^k.$

As a base step, let $x_1$ be some point in $R^k,$ and let $B_1$ be an open ball around $x_1.$ Since the interior of $F_1$ is empty, $B_1\cap F_1^c$ is not empty, and is open.

For the inductive step, suppose we have some $B_n$ that's constructed such that $B_n\cap F_n^c$ is non-empty and open. Then, we can pick a point $x_{n+1}\in B_n\cap F_n^c,x_{n+1}\ne x_n,$ and make a ball $B_{n+1}$ around it such that $B_{n+1}\subset B_n\cap F_n^c,$ $x_n\notin B_{n+1},$ and $B_{n+1}\cap F_{n+1}^c$ is open and not empty. Now, since $B_{n+1}$ satisfies our induction hypothesis, and since $B_1$ does as well, we have $B_n$ defined for all $n=1,2,3,\dots .$

Since the set of points $\{x_n\}$ is infinite (by induction) and bounded (all points are within $B_1$,) it has a limit point $p$ in $R^k.$. Now, suppose, for contradiction, that $p$ is not in every $F_n^c.$ Then, for some $N,$ $p$ is not in $F_N^c.$ Since $p\notin F_N^c$ and $B_{N+1}\subseteq F_N^c$, we have $p\notin B_{N+1}$. Since $B_{N+1}$ is open and $p$ is outside it, $p$ is at some positive distance $\epsilon$ from any point $q\in B_{N+1}$. But $x_n\in B_n\subseteq B_{N+1}$ for all , so all these infinitely many points are at distance at least $\epsilon$ from $p$. Thus any neighborhood of $p$ with radius less than $\epsilon$ contains at most finitely many points of $\{x_n\}$, contradicting that $p$ is a limit point.

Now, this means that $p\in \bigcap _{n=1}^{\mathrm{\infty }}{F}_n^c,$ meaning $\bigcap _{n=1}^{\mathrm{\infty }}{F}_n^c\ne \mathrm{\varnothing },$ a contradiction! Therefore, our supposition that every $F_n$ has an empty interior must be incorrect, and some $F_n$ must have a non-empty interior.