I had to answer three questions about accumulation points. I think my work is correct, but I'd appreciate if someone would look over them for me. (I'm not sure if I read question 2 correctly.)
In my proofs, I will define $x$ as an accumulation point of $S \subseteq \mathbb{R}$ if the defining condition holds: $\forall \epsilon > 0, \exists y \in S$ s.t. $y \neq x$ and $y \in (x-\epsilon,x+\epsilon)$.
(1)
Find an infinite subset of $\mathbb{R}$ that does not have an accumulation point in $\mathbb{R}$.
Take $\mathbb{Z}$, an infinite subset of $\mathbb{R}$. We can show that $\nexists x \in \mathbb{R}$ to satisfy the defining condition. Suppose there were some $x$ s.t. $\forall \epsilon > 0, \exists y \in \mathbb{Z}$ s.t. $y \neq x$ and $y \in (x-\epsilon, x+\epsilon)$, iff $|y-x| < \epsilon$. Note that as $y \neq x$, $y-x \neq 0$, so $0<|y-x|<\epsilon$. Then let $\epsilon = \frac{|y-x|}{2}$, so $0 < \epsilon < |y-x|$. So we can conceive an $\epsilon$ s.t. $y \notin (x – \epsilon, x+\epsilon)$, so the defining condition doesn't hold.
(2)
Find a bounded subset of $\mathbb{R}$ that does not have an accumulation point in $\mathbb{R}$.
Take $X = \{1,2,3\}$, so $X \subseteq \mathbb{Z} \subseteq \mathbb{R}$. See (1).
(3)
Find all the accumulation points of $\mathbb{R} \backslash \mathbb{Q}$ in $\mathbb{R}$. Note that $\mathbb{R} \backslash \mathbb{Q} = \mathbb{I}$, the irrationals.
The rationals and the irrationals are both dense in $\mathbb{R}$, so it follows that both $(\mathbb{I} \cap (x-\epsilon,x+\epsilon))\backslash\{x\} \neq \emptyset$ and $(\mathbb{Q} \cap (x-\epsilon,x+\epsilon))\backslash\{x\} \neq \emptyset$ hold for any $\epsilon >0$. We know this because all rationals and irrationals are reals, so any such rational or irrational $\pm \epsilon$ is also a real, so there exists both a rational and an irrational between any two reals $x-\epsilon, x+\epsilon$. Therefore, no matter how small we make our $\epsilon$, we will still have both a rational and an irrational in $(x-\epsilon,x+\epsilon)$. As the reals are the union of the disjoint sets $\mathbb{Q},\mathbb{I}$, it must be that $\mathbb{R}$ is the set of accumulation points for the irrationals.
Best Answer
As Dave L. Renfro said, the sentence "We know this because all rationals and irrationals are reals" is strange. We work here in the context of real line: there is nothing but real numbers, the real line is our Universe. Rational and irrational numbers were defined within this Universe, so saying they belong to it is redundant.
All you really need to do there is to argue that for every $x\in\mathbb R$ and every $\epsilon>0$ the interval $(x-\epsilon,x+\epsilon)$ contains an irrational number. This can be done by a cardinality argument: the interval is an uncountable set, while rational numbers are countable. There are more explicit ways too, like considering the set $\{ \frac{m}{n}\sqrt{2}:m\in\mathbb Z\}$ where $n$ is large enough so that $\sqrt{2}/n<\epsilon$. This set has to have a point in the interval.