Determine whether the given points are limit points of the given set Let $P = \{(x,y) \in \Bbb R^2 : 0 \le x^2 < 1, 2 < y \le 4 \}$ and $S = \{(x,y) \in \Bbb R^2 : 1\le x \le 3, y = 2 \}$.

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*Are $(0,3)$ and $(0,4)$ be the limit points of $P$?

*Are all points of $S \cup P$ be the limit points of $S \cup P$?

Attempt:
For 1, I tried on $(0,3)$ first like this.
Let $r > 0 $. To check: $(0,3)$ is a limit point of $P$. In other words, $B((0,3),r) \cap P \setminus \{(0,3) \} \ne \emptyset$.
Let $(x,y) \in B((0,3),r) \setminus \{(0,3)\}$. Then, we have
\begin{align*}
x^2 + (y-3)^2 < r^2 \\
x^2 < r^2 \ \wedge (y-3)^2 < r^2 \\
-r < x < r \ \wedge 3-r < y < 3+r
\end{align*}
Consider the interval $(3,3+r)$. We take a smallest $y$ in $(3,3+r)$, say $y = 3 + \frac{r}{3}$.
Then, we have
\begin{align*}
x^2 + ((3 + \frac{r}{3}) - 3)^2 = \frac{r^2}{9} < r^2
\end{align*}
which means that $(0,3+\frac{r}{3}) \in B((0,3),r) \setminus \{(0,3) \}$. On the other hand, it's clear that $(0,3+\frac{r}{3}) \in P$. Hence, we have $B((0,3),r) \cap P \setminus \{(0,3)\} = \{(0,3+\frac{r}{3}) \} \ne \emptyset$. Therefore, $(0,3)$ is a limit point of $P$.
Is it true? If yes, then the analog could be used on $(0,4)$.
For 2, I'm not sure that the answer is true since there exists point that would be outside of $S \cup P$. Here's what I tried.
I'm considering the point $(3,2) \in S \cup P$ and take $r = 1 > 0$.
To show: $B((3,2), 1) \cap (S \cup P) \setminus \{(3,2) \} = \emptyset$.
Let $(x,y) \in B((3,2),1) \setminus \{(3,2)\}$. Then, we have
\begin{align*}
(x-3)^2 + (y-2)^2 < 1 \\
-1 < x-3 < 1 \ \wedge -1 < y-2 < 1 \\
2 < x < 4 \ \wedge 1 < y < 3.
\end{align*}
But, the point $(3.5, 2) \notin S \cup P$. Hence, we have
$B((3,2),1) \cap (S \cup P) \setminus \{(3,2)\} = \emptyset$.
Am I true?
 A: The fact that $\lambda$ is a limit point of a set $S$ is also equivalent to the fact there exists a sequence $(x_n)_{n \in \mathbb{N}} \subseteq S \setminus \{\lambda\}$ such that $x_n \to \lambda.$ This is usually easier to check than the exact definition of $S'$ (they are in fact equivalent). For example, you might have noticed that $((\frac1n, 3 + \frac1n))_{n \in \mathbb{N}} \subseteq P \setminus \{(0, 3)\}$ and the limit of this sequence is of course equal to $(0, 3),$ so this is a limit point indeed. Analogously one shows that $(0, 4) \in P'.$
Your reasoning of the second part is not entirely sound. $(3, 2) \in (S \cup P)'$ because $((3 - \frac1n, 2))_{n \in \mathbb{N}} \subseteq (S \cup P) \setminus\{(3, 2)\}.$ It is a general fact (and also not hard to prove) that the closure of a set is the union of its limit and isolated points (disjoint union, of course). Note that $S$ is closed and has no isolated points. Therefore, any point of $S$ is a limit point of $S,$ so it is also a limit point of $S \cup P.$ The closure of $P$ is $\overline{P} = [-1, 1] \times [2, 4]$ and it clearly has no isolated points, so $P \subseteq \overline{P} = P',$ hence any point in $P$ is a limit point of $P,$ so also a limit point of $S \cup P.$ I hope this helps. :)
