Show that $\{(x,\sin(1/x)) : x∈(0,1] \} \cup \{(0,y) : y ∈ [-1,1] \}$ is closed in $\mathbb{R^2}$ using sequences Usual metric.
The abridged definitions we were just given are:
Convergent Sequence: A sequence $(x_n)$ in a metric space $X$ converges to $x_0$ if for every $\varepsilon \gt 0$ there is $n_0\in\mathbb{N}$ such that $d(x_n, x_0) \lt \varepsilon$ for each $n \geq n_0$.
Closed set: A set $F$ in a metric space $X$ is closed if and only if every convergent sequence in $F$ converges to a point in $F$.

This is the last question of a problem set to study for class. The definitions we are using are very standard for an Intro to Topology course, I just don't understand how to apply them properly yet. Would anyone be able to lend me a hand? Or show me how to start?
 A: Let a sequence $(x_n,\sin\frac{1}{x_n})$ converge to some point $(x_0,y_0)$. Since $x_n\ge 0$ and $-1\le y_n\le 1$, we have two cases:

*

*$x_0=0$
In this case, the limit point falls within $\{(0,y): -1\le y\le 1\}$ because $x_n\ge 0$ and $-1\le y_n\le 1$.

*

*$x_0>0$
We can write
$$
{
|x_n-x_0|<\epsilon_1\implies |\sin\frac{1}{x_n}-\sin\frac{1}{x_0}|<\epsilon_2
\\
|\sin\frac{1}{x_n}-y_0|<\epsilon_3
}.
$$
From the above inequalities, we conclude by the triangle inequality that
$$
{|\sin\frac{1}{x_0}-y_0|
\\=|\sin\frac{1}{x_n}-y_0-\sin\frac{1}{x_n}+\sin\frac{1}{x_0}|
\\\le|\sin\frac{1}{x_n}-y_0|+|\sin\frac{1}{x_n}-y_0|
\\<\epsilon_2+\epsilon_3
}
,
$$
which yields $\sin\frac{1}{x_0}-y_0$ because both $y_0$ and $\sin\frac{1}{x_0}$ are constant $\blacksquare$
A: Let $S$ denote the set in question and fix $z \in S$.
First suppose that $z = (x,\sin(\frac{1}{x}))$ for some $x \in (0,1]$. Pick your favourite sequence $(x_n)$ such that $x_n \to x$ and $x_n \in (0,1]$ (eg. $x_n = x + \frac{1}{n}$ for large enough $n$.) ​Since $\sin(\frac{1}{x})$ is continuous on $(0,1]$ it follows that $\sin(\frac{1}{x_n}) \to \sin(\frac{1}{x})$. Hence, $(x_n,\sin(\frac{1}{x_n}))$ is a sequence in $S$ which converges to $z$.
Now suppose that $z = (0,y)$ for some $y \in [-1,1]$. Since the function $\sin$ is continuous, an Intermediate Value Theorem argument implies that for each $n \in \mathbb N$, there exists some $a_n \in [n,n+2\pi]$ such that $\sin(a_n) = y$. Let $x_n = \frac{1}{a_n}$ and observe that $x_n \in (0,1]$. Then $x_n \to 0$ and $\sin(\frac{1}{x_n}) \to y$. It follows that $(x_n,\sin(\frac{1}{x_n}))$ is a sequence in $S$ converging to $z$.
Since every point of $S$ is the limit point of some sequence in $S$, it follows that $S$ is closed.
A: Let $A: = \{ (x, \sin(\frac{1}{x})) \mid x \in (0, 1] \}$ and $B := \{ (0, y) \mid y \in [-1, 1] \}$.
Fix any sequence $\{a_n\} _{n \in \mathbb{N}} = \{ (x_n , y_n )\} _{n \in \mathbb{N}}$ in $A \cup B$ that converge in $\mathbb{R}^2$.
Let $a = (x,y)$ be the limit point of $\{a_n\}_n$.
Let $N_A := \{ n \in \mathbb{N} \mid a_n \in A \} $ and $N_B := \{ n \in \mathbb{N} \mid a_n \in B \}$.
Then, at least one of $N_A$ and $N_B$ is unbounded in $\mathbb{N}$.
Let us assume that the essential case  that $N_A$ is unbounded (otherwise use closeness of $B$) and $x = 0$ (otherwise use continuity of $\sin(\frac{1}{x})$) occurs.
Consider the subsequence $\{a_n\}_{n\in N_A}$.
Note that the limit of any subsequence of a convergent sequence $\{p_n\}_{n \in \mathbb{N}}$ is equal to the limit of $\{p_n\}_{n \in \mathbb{N}}$ (you know?).
Since $y_n =  \sin (\frac{1}{x_n})\in [-1 , 1]$ for all $n \in N_A$ and $[-1, 1]$ is closed, we have $y \in [-1, 1]$.
Therefore $a \in B$.
Thus $A\cup B$ is closed under limit of sequence and hence it is closed.
