$Show that ${(x,\sin(1/x)) : x∈(0,1]} ∪ {(0,y) : y ∈ [-1,1]}$ is closed in $\mathbb{R^2}$ using sequences$

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$Show that ${(x,\sin(1/x)) : x∈(0,1]} ∪ {(0,y) : y ∈ [-1,1]}$ is closed in $\mathbb{R^2}$ using sequences$

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Convergent Sequence: A sequence $(x_n)$ in a metric space $X$ converges to $x_0$ if for every $\varepsilon>0$ there is $n_0\in\mathbb{N}$ such that $d(x_n, x_0)<\varepsilon$ for each $n≥n_0$.

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 $F$.

Usual metric.

Topology Sequences and Series

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Let $(a_n,b_n)$ be a sequence in the set converging to $(a,b)$. We need to show $(a,b)$ is also in the set. Since for every point $(x,y)$ in the set we have $|y|\le 1$, we have $|b_n| \le 1$, so $|b|\le 1$ (because $b_n\to b$). Now if $a=0$ then $(a,b)$ is in the set.

So suppose $a\ne 0$. Then we have $a_n \to a$ so for large enough $n$ the $a_n$ is also nonzero. So we must have $$(a_n,b_n)=(a_n, \sin (1/a_n))$$ But $\sin(1/x)$ is continuous at nonzero $x$, so $\sin(1/a_n)\to \sin (1/a)$, and we get $$(a,b) =\lim (a_n,b_n)= \lim (a_n, \sin (1/a_n)) =(a, \sin (1/a))$$ So $(a,b)$ is in the set in this case too.

Martin

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$Show that ${(x,\sin(1/x)) : x∈(0,1]} ∪ {(0,y) : y ∈ [-1,1]}$ is closed in $\mathbb{R^2}$ using sequences$

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