3

Inspired on proving that every compact set of the Sorgenfrey line is countable.

Trying to prove any of these in $\mathbb{R}$: 1) Every bounded non countable set has a both-sided accumulation point. That is, I supposed a stronger version of Bolzano's theorem.

2) Every bounded non countable set, has an interval of irrational numbers as subset. And I think its almost the same as saying that the bounded-non-countable set has a subset where every point is an accumulation point.

First of all they might not be true(thou I fell very sure they are), and second I dont have yet the tools to prove it even knowing that its not that complicated.

DavidBecharaSenior
  • 201
  • What is a 'both-sided' accumulation point (how about $0$ in $(0,1)$)? And what is an 'interval of irrational numbers'? – copper.hat Jun 06 '13 at 01:29
  • Im new so I'm having a hard time writing everything. – DavidBecharaSenior Jun 06 '13 at 01:47
  • In $(0,1)$ there exists $0,5$ which has the needed property, by both-sided accumulation point I mean that you can find a point $p\in S$ such that, $\forall \epsilon, (p−\epsilon,p)\cap S\ne \phi,and,(p,p+\epsilon)\cap S\ne \phi$. By "interval of irrational numbers" i mean this $[a,b]\cap \mathbb{I},a,b\in mathbb{R}$ – DavidBecharaSenior Jun 06 '13 at 02:01
  • 1
    The word "both-sided" actually exists in English, but the word you want in this context is "two-sided". By the way, you don't need "bounded" in your question; every uncountable set of real numbers contains an uncountable bounded subset. – bof Sep 13 '22 at 06:49

3 Answers3

5

(1) Let $X\subseteq\Bbb R$ be uncountable. For $n\in\Bbb Z^+$ let $$L_n=\left\{x\in X:\left(x-\frac1n,x\right]\cap X\text{ is countable}\right\}$$ and $$R_n=\left\{x\in X:\left[x,x+\frac1n\right)\cap X\text{ is countable}\right\}\;.$$

Fix $n\in\Bbb Z^+$. The family $\left\{\left[x,x+\frac1n\right):x\in R_n\right\}$ is an open cover of $R_n$ in the Sorgenfrey topology, which is hereditarily Lindelöf, so there is a countable $C\subseteq R_n$ such that $\left\{\left[x,x+\frac1n\right):x\in C\right\}$ covers $R_n$. But for each $x\in C$ we have $\left[x,x+\frac1n\right)\cap R_n\subseteq\left[x,x+\frac1n\right)\cap X$, a countable set, so $R_n$ is countable. The same argument using the reverse Sorgenfrey topology shows that each $L_n$ is countable. Thus,

$$\bigcup_{n\in\Bbb Z^+}(L_n\cup R_n)$$

is countable. Let $$X_0=X\setminus\bigcup_{n\in\Bbb Z^+}(L_n\cup R_n)\;;$$

you can check that for each $x\in X_0$ and each $\epsilon>0$, the sets $[x,x+\epsilon)\cap X_0$ and $(x-\epsilon,x]\cap X_0$ are not just non-empty, but actually uncountable.

(2) is false as stated. Let $\mathscr{B}$ be the set of open intervals with rational endpoints; $\mathscr{B}$ is countable, so we can enumerate it as $\mathscr{B}=\{B_n:n\in\Bbb N\}$. For each $n\in\Bbb N$ let $x_n\in B_n$ be irrational. Let $D=\{x_n:n\in\Bbb N$, and let $X=[0,1]\setminus D$; certainly $X$ is uncountable. Suppose that $[a,b]\cap X$ is an interval in $X$; without loss of generality we may assume that $0\le a<b\le 1$. Then there is an $n\in\Bbb N$ such $B_n\subseteq[a,b]$, and by construction $x_n$ is then an irrational in $[a,b]\setminus X$. Thus, $X$ contains no interval $[a,b]\setminus\Bbb Q$ of irrationals.

Community
  • 1
Brian M. Scott
  • 616,228
  • Just to make it explicit, the answer shows: Every uncountable subset of $\mathbb{R}$ has a co-countable number of two-sided condensation points. – PatrickR Oct 16 '21 at 03:16
2

Rudin - Principals of Mathematical Analysis - Page 45 - Problem 27

All but a countable number of points are accumulation points in every uncountable subset of $\mathbb{R}^n$.

In fact, all but a countable number of points are condensation points in every uncountable subset of $\mathbb{R}^n$. A condensation point of a set $E$, is a point in which every neighborhood contains uncountably many points of $E$. Obviously, if a point is a condensation point then it is an accumulation point.

Not sure exactly what you mean by "an interval of irrational numbers". However, it should be noted that it is entirely possible to define a subset of $\mathbb{R}$ that is uncountable and has no rational number as an accumulation point.

Bobby Ocean
  • 3,205
0

Consider $(X, \tau) $ a topological space, and $A$ a subset of $X$.

Let $$B \colon = \{ b \in A \ | \ \exists U \in \tau, b \in U \ \ \& \ U \cap A \textrm{ countable } \}$$

Then (almost a tautology) $$B = \bigcup_{U\in \tau,\ U \cap A \ \textrm{countable}} (U \cap A)$$

Assume now that $X$ is a Lindelöf space ( any topological space with a countable basis, like $\mathbb{R}^n$ will do). Then the family of open subsets $\{ U \in \tau \ | \ U \cap A \ \textrm{countable} \}$ can be replaced with a countable subfamily that still covers $B$. We conclude that $B$ is countable.

If moreover, $A$ is uncountable, we get that the cardinality of $A\backslash B$ equals the cardinality of $A$.

Conclusion: if $X$ has a countable basis, and $A\subset X$ is uncountable, then there exists and uncountable subset $A_1\subset A$ such that for every $a \in A_1$, and $U$ neighborhood of $a$, we have $U\cap A$ is uncountable.

orangeskid
  • 53,909