I was recently going through a proof of the result that:
"The set $\Bbb R$ is uncountable"
The proof I encountered was:
This result might seem unexpected, but its proof can be done by contradiction. Assume that there does exist a $1–1,$
onto function $f : \Bbb N → \Bbb R.$ Again, what this suggests is that it is possible to enumerate the elements of $\Bbb R.$ If we let $x_1 = f(1), x_2 = f(2),$ and so on, then our assumption that f is onto means that we can write
$$\Bbb R = \{x_1, x_2, x_3, x_4,… \}\tag 1$$
and be confident that every real number appears somewhere on the list. We will now use the Nested Interval Property to produce a real number that is not there. 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. Certainly $I_1$ contains two smaller disjoint closed intervals, and $x_2$ can only be in one of these. In general, given an interval $I_n$, construct $I_{n+1}$ to satisfy:
(i) $I_{n+1}\subseteq I_n$ and
(ii) $x_{n+1 }\notin I_{n+1}$
We now consider the intersection $\bigcap_{n=1}^{\infty}I_n.$ If $x_{n_0}$ is some real number from the list in $(1),$ then we have $x_{n_0}\notin I_{n_0}$ , and it follows that $x_{n_0}\notin\bigcap_{n=1}^{\infty}I_n.$ Now, we are assuming that the list in $(1)$ contains every real number, and this leads to the conclusion that $\bigcap_{n=1}^{\infty}I_n=\emptyset.$ However, the Nested Interval Property (NIP) asserts that $\bigcap_{n=1}^{\infty}I_n\neq\emptyset.$ By NIP, there is at least one $x ∈ \bigcap_{n=1}^{\infty}I_n,$ that consequently, cannot be on the list in $(1).$ This contradiction means that such an enumeration of $\Bbb R$ is impossible, and we conclude that $\Bbb R$ is an uncountable set.
However, I am having problem with two lines of this prooof. They are:
(i) The existence of such an $I_2$ is easy to verify.
(ii) Certainly $I_1$ contains two smaller disjoint closed intervals, and $x_2$ can only be in one of these.
For the line (i), I don't understand what are they trying to implying by the phrase "easy to verify" in this context.
The existence of such an $I_2$ looks trivial to me. For if, $I_1=[a,b]$, and if $x_2\lt a$ or $x_2\gt b,$ then any interval $J\subset I_1$ can be called $I_2,$ (i.e we can then call $J=I_2$). If this is not the case, then $x_2\in I_1,$ so, we can easily choose, an $a_1$ such that $a_1=\frac{b+x_2}{2}$ and say, $b_1=\frac{a_1+b}{2},$ and say, that $I_2=[a_1,b_1]\subset I_1,$ such that $x_2\notin I_2.$ I find this lines of rasoning is more than adequate to justify the existence of such an $I_2.$
Is this what the author meant by the phrase , "easy to verify" ?
Next, I am having some problem, with my point number (ii) i.e , "Certainly $I_1$ contains two smaller disjoint closed intervals, and $x_2$ can only be in one of these". Now, this is absurd and ambiguous. This is because, it is very much evident, that $I_1$ can be divided into two disjoint closed subintervals, but I don't understand, which two subintervals are being talked about here. Incidentally, it might very well be the case, that $I_2$ divides the interval $I_1$ into three intervals (counting $I_2$). So, the phrase "Certainly $I_1$ contains two smaller disjoint closed intervals" makes no sense to me.
Now, I am focussing on the remaining part of the sentence, i.e "and $x_2$ can only be in one of these". This claim also seems erroneous to me, for it might happen that $x_2\lt a$ or $x_2\gt b$ and so, $x_2\notin I_1$ and thus, $x_2$ should not be in any interval (open or closed) inside $I_2.$
I want to know precisely, whether I am justified in my statements or not?
Best Answer
Easy to verify means easy to verify. You verified the statement easily, so... not sure what the problem is there.
What they're saying is neither absurd nor ambiguous nor erroneous. They simply suggested a way in which you might verify the claim. You've already verified it for yourself, so those sentences don't pertain to you.
If $I_1=[a,b]$, then you can pick two, strictly smaller in diameter, closed subintervals $I,I'$ of $I$ which are disjoint. There are myriad (uncountably many, technically) ways to do this and the precise detail is completely unimportant. If you insist, we can make it precise by choosing $I=[a,a+\frac{b-a}{4}]$ and $I'=[b-\frac{b-a}{4},b]$.
"$x_2$ can only be in one of these": this is true. It might be in neither of them, but that wouldn't contradict what they said ("can", not "is"). Admittedly it would have been sharper to say "$x_2$ can be in at most one of these", but it's fine. Because of this, we know $x_2\notin I$ or $x_2\notin I'$ will be true, so we can select $I_2$ appropriately from the options $I,I'$.