Mathematics III - Foundations of Analysis - Summer 24

Hi!
Let me try to break down the answer to the different parts of your question here.

Q: "what prevents any arbitrary a + 1/n from being considered a limit point if n can be any arbitrarily large natural number (since a point p is a limit point if every neighbor of p has at least one point in E)?"

A: Do you mean that $p = a + \frac{1}{n}$ should be a limit point of the set since every neighbourhood $U$ of $p$ has a point in $E$, namely $p$ itself. If so this is a very good observation but unfortunately your definition of a limit point is slightly incorrect. The correct definition is that $p$ is a limit point of a set $S$ if every neighbourhood $U$ of $p$ contains at least one point in $S$ EXCEPT $p$ itself. Using mathematical notation we say that $p$ is a limit point of $S$ if
$S \cap (U \setminus p) \neq \emptyset.$
This means that $p= a+\frac{1}{n}$ is not a limit point of $E$ since the open neighbourhood $(a +\frac{1}{n+1}, a+\frac{1}{n-1})$ of $p$ contains no points in $E$ except for $p$ itself.

I am not sure if my interpretation of this question is wrong so let me know if this was not the answer you were looking for.

Q: "what is "The set U of y" meant to be? What is y in that context?"

A: I agree that the formulation in the solution is a bit unclear. What they mean is that $U$ is the set of all real numbers $y$ such that $|x-y| < \frac{\delta}{2}$. That is $y$ is just used to denote a point in $U$, or perhaps more accurately, $y$ denotes any point and then they give a condition for which points $y$ are points in $U$. Using mathematical notation we would write
$U = \{y \in \mathbb R\ | \ |x-y| < \frac{\delta}{2} \} .$
When $W$ is defined $y$ is used in the same way. It denotes a point and then they give a condition for when such a point is an element of $W$.

I hope that helps but if not then please let me know and I will try to clarify!