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This rabbit hole led me to try proving $[0,1)$ a half closed-interval is uncountable via the Nested Interval property.

I got distracted proving the bijection $(0,1)$ ~ $[0,1)$ and tried to prove both intervals are uncountable, and so bijective with the real numbers, and so bijective with each other.

But, I'm just trying to confirm my proof that HNIP for $[0,1)$ proves uncountability.

This is not a repeat question. I have looked:

But I have proven it (possibly) for $[0,1)$ and possibly more generally for $[a_n,b_n)$ with the element $x \ne b_n$ giving non-empty intersection.

If I've made a mistake, please be nice. I've spent too long on this. Hopefully my motivational context is sufficient for MSE standards, and my question is clear enough to answer.

Half-closed nested interval property

Here is my proof that: the nested interval property $\bigcap_{n=1}^{\infty} I_n \ne \emptyset$ holds for half-closed nested intervals $ I_n = [a_n, b_n) = \{x \in \mathbb{R} : a_n \leq x < b_n\} $ so long as $ x \in \bigcap_{n=1}^{\infty} I_n$ is not some right-hand endpoint $x=b_n$, with the example of $[0,1)$.

First, look at arbitrary half-closed nested intervals. Say we have some half-closed interval $I_n = [a_n,b_n)$. There is some nested interval $I_{n+1} = [a_{n+1},b_{n+1})$ satisfying $I_{n+1} \subseteq I_{n}$. (So we found some $a_{n+1}$ and $b_{n+1}$ where $a_n \le a_{n+1}$ and $b_{n+1} < b_n$).

Specifically, for $n=1$, say $I_1 = [0,1)$. To see the non-emptiness of the intersection $\bigcap_{n=1}^{\infty} I_n \ne \emptyset$, we use the Axiom of Completeness (AoC) to find a real number $x$ such that $x \in I_n$ for all $n \in \mathbb{N}$ where $I_1 = [0,1)$. So, informally $x \in I_{\infty} \subseteq ... \subseteq ... \subseteq I_{n+1} \subseteq I_n \subseteq ... \subseteq I_3 \subseteq I_2 \subseteq I_1$.

To see conditions for AoC, say the sets $A=\{a_n:n \in \mathbb{N}\}$ and $B=\{b_n:n \in \mathbb{N}\}$ are all the left-hand and right-hand endpoints of the intervals, respectively.

Given the intervals $... \subseteq I_{n+1} \subseteq I_n \subseteq ... \subseteq I_3 \subseteq I_2 \subseteq I_1 \subseteq$ are nested, each right end point $b_n$ serves as an upper bound for all left end points $A$. So, by AoC, we see $A$ has at least one upper bound and so must have a "least" upper bound—or supremum.

So say $x=\sup A$, (where $A$ is the set of left-hand endpoints).

As $x$ is an upper bound for $A$, we have $a_n \leq x$. Also, since each $b_n$ is an upper bound for $A$ and $x$ is the least upper bound, we see $x \leq b_n$.

So, we establish $a_n \leq x \leq b_n$. But since each $I_n=[a_n,b_n)={x:a_n \leq x < b_n}$, we go one step further.

First see: $x \in [a_n,b_n]$ does not give $ x\in [a_n,b_n)$ since $x=b_n \in [a_n,b_n]$ while $x=b_n \notin [a_n,b_n)$.

It is sufficient to see $a_n \leq x \leq b_n$ implies $a_n \leq x < b_n$ for $x \ne b_n$ since $[a_n,b_n) = [a_n,b_n] /\ \{b_n \} \Rightarrow [a_n,b_n) \subseteq [a_n,b_n]$, giving $x \in I_n$ for all $n \in \mathbb{N}$.

Note: $x \in [a_n,b_n]$ does not give $ x\in [a_n,b_n)$ for $x=b_n$

Thus $x \in \bigcap_{n=1}^{\infty} I_n$ and so $\bigcap_{n=1}^{\infty} I_n \ne \emptyset$ since for all $x$ we know $x \notin \emptyset$. □

For the case of $[0,1)$ one way to see this is by defining $I_n=\{x:0 \le x < \frac{n}{n+1}\}$ giving $\bigcap_{n=1}^{\infty} I_n = [0,\frac{1}{2})$. and since $[0,\frac{1}{2}) \ne \emptyset$ we see $\bigcap_{n=1}^{\infty} I_n \ne \emptyset$

See: $0 \leq x < \frac{n}{n+1}$ and $\frac{1}{2} \leq \frac{n}{n+1} < 1$ gives $0 \leq x < \frac{1}{2}$

Uncountablity follows from HNIP

This uses the same reasoning as Abbott's proof for the uncountability of the real numbers.

Say there exists a one-to-one, onto function $ f : \mathbb{N} \to [0,1) $, giving the possibility of enumerating the elements of $[0,1)$. If we say $ x_1 = f(1) $, $ x_2 = f(2) $, and so on, our assumption that $ f $ is onto allows us to write $[0,1) = \{x_1, x_2, x_3, x_4, \ldots\}$ where every real number in $[0,1)$ appears somewhere on the list. We will now use the Half-closed Nested Interval Property to produce a real number in $[0,1)$ that is not included in this list.

Let $ I_1 $ be a closed interval that does not contain $ x_1 $. Next, let $ I_2 $ be a closed interval, contained in $ I_1 $, which does not contain $ x_2 $. The existence of such an $ I_2 $ is easy to verify. See, $ I_1 $ contains two smaller disjoint closed intervals, and $ x_2 $ can only be in one of these.

Note: the property above follows from the partitioning of supersets by proper subsets. This is explained here: https://math.stackexchange.com/a/2724109/1098426

In general, given an interval $ I_n $, construct $ I_{n+1} $ to satisfy (i) $ I_{n+1} \subseteq I_n $ and (ii) $ x_{n+1} \notin I_{n+1} $.

Now, consider the intersection $\bigcap_{n=1}^{\infty} I_n$. If $ x_{n_0} $ is some real number in $[0,1)$ from the list $[0,1) = \{x_1, x_2, x_3, x_4, \ldots\}$, then we have $ x_{n_0} \notin I_{n_0} $, and it follows that $ x_{n_0} \notin \bigcap_{n=1}^{\infty} I_n $.

We are assuming that the list $[0,1) = \{x_1, x_2, x_3, x_4, \ldots\}$ contains every real number in $[0,1)$, leading to the conclusion that $\bigcap_{n=1}^{\infty} I_n = \emptyset$. However, the Half-closed Nested Interval Property (HNIP) asserts that the intersection is non-empty, so $\bigcap_{n=1}^{\infty} I_n \ne \emptyset$. By HNIP, there is at least one $ x $ in $\bigcap_{n=1}^{\infty} I_n$ that cannot be on the list $[0,1) = \{x_1, x_2, x_3, x_4, \ldots\}$. This contradiction means that such an enumeration of $[0,1)$ is impossible, and we conclude that $[0,1)$ is an uncountable set. □

isaac
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1 Answers1

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In short:

$(1)$ Density of $\mathbb{R}$ in $\mathbb{R}$ ($\forall a,b \in \mathbb{R} : \exists x \in \mathbb{R} : a<x<b$) $\quad \Rightarrow$

$(2)$ Nested interval property for half-closed intervals $\exists x \in \mathbb{R} : x \in \bigcap_{n=1}^\infty [a_n,b_n)$ where $[a_{n+1},b_{n+1})\subseteq[a_n,b_n)$ and $a_n < b_n$ $(*)$ $\quad \Rightarrow$

$(3)$ Uncountability of half-closed intervals $[a,b) \subset \mathbb{R}$

$(*)$ The existence of some $b_n : b_n > a_n \forall n \in \mathbb{N}$ follows from fixing $a_n$ and identifying $b_n$ with the infinum for (the set of points of) a series with accumulation points on the open interval $(a_n, b_n)$.

See details below

In words: my half-closed nested interval property (NIP) formulation & proof reduces to the density of $\mathbb{R}$ in $\mathbb{R}$, (or that $\mathbb{R}$ has no isolated points).

Then, the uncountability of half-closed intervals (following this nested interval theorem) follows from the partition of a superset into at least two disjoint proper subsets for every proper subset. For countably many nested proper subsets as in (NIP), we can always find some $x$ in the other nested subsets created by the partition.

In my proof of the Nested Interval Property (NIP) for half-closed intervals, I required that we find some $x$ with two conditions. The first is $(1)$ that $x \in \bigcap_{n=1}^\infty [a_n,b_n)$ as in the traditional proof of the NIP for closed interval$[a_n,b_n]$. I also introduced the requirement $(2)$ that this $x$ satisfies $x\ne b_n$ to avoid the issue of $[x,x)$.

This second condition $(2)$ is better stated as $a_n<b_n$ so that $[a_n,b_n) \ne [b_n,b_n) \ne [a_n,a_n)$, which is not part of the traditional closed NIP. In fact, the general traditional case often has $x \in \bigcap_{n=1}^\infty [a_n,b_n]$ following from $\bigcap_{n=1}^\infty [a_n,b_n]=[x,x]=\{x\}$ since $x\in\{x\}$.

In my example of $[0,1)$, I showed We can find some $a_n \le x <b_n$ given $a_n<b_n$ with $n\in\mathbb{N}$ by fixing $a_n$ and counting $b_n$ as the infinum of the set given by some series on $[a_n,b_n)$. That is, $b_n = \inf(S)$ for $S= \{x:x=s_n : (s_n)_{n=1}^\infty\}$ where $\{x:x=s_n : (s_n)_{n=1}^\infty\} \subset (a_n,b_n)$. Of course, the case $x=a_n$ is given too.

So, the statement $\bigcap_{n=1}^\infty [a_n,b_n) \ne \emptyset$ such that each $[a_{n+1},b_{n+1})\subseteq[a_n,b_n)$ and such that reduces to the density of the reals in the reals.

It looks like the denisty of $\mathbb{R}$ in $\mathbb{R}$ is a simply property based on this answer: https://math.stackexchange.com/a/3378022/1098426

I can also prove it by saying: $a<b$ gives $0<b-a$. Then, say $b-a=\epsilon$ gives $0<\epsilon$. Then, say $\alpha<\epsilon$ so $b-\alpha > b-\epsilon$. But, $b-\epsilon=a$ so $b-\alpha>a$. We know $b-\alpha<b$. So, with the final two ineqalities we see $a<b-\alpha<b$. Say $x=b-\alpha$ so there is some $a<x<b$ given $a<b$.

Since the half-closed NIP holds (again, following the denisty of $\mathbb{R}$ in $\mathbb{R}$), the uncountability of half-closed intervals following the half-closed NIP also holds.

isaac
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