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Show $\mathbb{R}$ is of 2nd category, ie. not a union of countably many nowhere dense sets using nested interval theorem.

Suppose, by way of contradiction, $\exists$ countably many no where dense sets $S_n$ such that $\displaystyle{\mathbb{R}\bigcup_{n=1}^{\infty}S_n}.$ Then, $\exists a_1<b_1\ s.t.\ [a_1,b_1]\cap S_1=\varnothing$.

Since we are trying to prove by contradiction that $\mathbb{R}$ is nowhere dense, we are trying to show that it is of 1st category, thus we show that it satisfies the definition of no where dense sets. In order to show that $a_n\leq b_1$ and therefore $A=\{a_n,n\in\mathbb{N}\}$ is bounded above by completeness and there exists a unique single point x, i.e. $\exists \sup A=x$.

Next, $\exists\ a_2<b_2\ s.t.\ [a_2,b_2]\subset[a_1,b_1],\ b_2-a_2\leq\displaystyle{\frac{b_1-a_1}{2}},\ and\ [a_2,b_2]\cap S_2=\varnothing$.

Using nested interval theorem we split our closed intervals into two subinterval halves, i.e. $\displaystyle{\left[a_1,\frac{a_1+b_1}{2}\right]}$ and $\displaystyle{ \left[\frac{a_1+b_1}{2},b_1\right]}$ knowing that one of the halves contains infinitely many elements. Knowing that subset of nowhere dense set is also nowhere dense meaning we need to show that there exists a subset interval where its intersection with $S_n$ is empty.

Similary, $\exists a_3<b_3\ s.t.\ [a_3,b_3]\subset [a_2,b_2],\ b_3-a_3\leq\displaystyle{\frac{b_2-a_2}{2}},\ and\ [a_3,b_3]\cap S_3=\varnothing$. Etc.\ Continue in this manner to obtain a sequence: $$[a_1,b_1]\supset[a_2,b_2]\supset[a_3,b_3]\supset ...$$ Then $\exists x\in\displaystyle{\bigcap_{n=1}^\infty}[a_n,b_n].$

We have continued dividing and relabeling sets by induction to obtain a collection of closed intervals so that

$$ [a_n,b_n]\supset[a_{n+1},b_{n+1}], \forall n\in\mathbb{N}. $$

And therefore by induction process,

$$ b_n-a_n=\frac{b_{n-1}-a_{n-1}}{2}=\frac{b_1-a_1}{2^{n-1}}, \forall n\geq 2 $$

Because we want to show that $\exists!x$ and therefore $\mathbb{R}$ has a single unique point and thus is nowhere dense. We may only accomplish this by showing a closed bounded set in $\mathbb{R}$ knowing that it is not possible as there will be infinitely many open non-empty dense sets in $\mathbb{R}$ that are uncountable and therefore not nowhere dense by contradiction, since $\mathbb{R}$ does not have a supremum.

$\forall n,x\not\in S_n.$

Suppose by contradiction, $\exists \sup A=x$, so $a_n\leq x\leq b_n, \forall n\in\mathbb{N}.$ Then $\exists n_0\ s.t.\ x\in[a_n,b_n]$. By downstream property then, $\exists y\in S\ s.t.\ \displaystyle{b_{n_0} +\frac{\epsilon}{2}\leq y\leq x}$ which is also a subset of A, but this contradicts that $x=\sup S$, therefore $\displaystyle{x\not\in\bigcap_{n=1}^\infty}[a_n,b_n].$

Since real numbers have infinitely many open non-empty sets, and sets such as $\mathbb{Q}$ we know that nested interval theorem is bound to fail in real numbers because NIT shows that the infinite intersection becomes the singleton for a closed interval, but this is not the case for open intervals.

Hence, $x\not\in\mathbb{R}$

This is because real numbers are closed and open at the same time, meaning that it contains its' interior and limit points. Which is a direct conclusion that $\sup\mathbb{R}=\infty.$ Hence nested interval theorem only works for closed sets, and that $\mathbb{R}$ is dense everywhere because its closure is $\mathbb{R}$.

I am new to the subject so I would appreciate some critique.. Thank you.

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I feel like this has been over complicated. You are all good defining $\mathbb{R} = \cup_{i=1}^\infty S_i$ and noting that by nowhere density $\mathbb{R} \setminus S_1$ contains a closed interval $[a_1, b_1]$. We now note that $S_2 \cap [a_1, b_1]$ is nowhere dense in $[a_1, b_1]$ and so can choose $$[a_2,b_2] \subset S_2 \cap [a_1,b_1].$$ By repeating this we define a sequence $$[a_1, b_1] \supset [a_2, b_2] \supset \ldots$$ where $$S_n \cap [a_n, b_n] = \emptyset$$ and so by the nested interval theorem there exists $x \in \cap_{i=1}^\infty [a_i,b_i]$. If we choose any $S_n$ then $x \in [a_n, b_n]$ and so $x \notin S_n$. This gives us that $x \notin \cup_{i=1}^\infty S_i $ and thus $\mathbb{R} \neq \cup_{i=1}^\infty S_i$.

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  • $\begingroup$ Thanks I do tend to complicate stuff. $\endgroup$ – C. Ekinci Apr 13 '18 at 0:02

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