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The problem that I'm having trouble with is this. \begin{align*} A := \bigcup_{n=1}^{\infty} B_{\frac{1}{n}}\!\!\left(\frac{1}{n},n\right). \end{align*}

Determine the closure of this subset.

By drawing the graph in $\mathbb{R}^2$, I was able to figure out that all these open balls are in contact with the y-axis, and the union of all these balls(=$A$) somehow looks like a cone in the Euclidean space. But I'm not sure how I can express the closure of $A$. To do this, I should come up with the accumulation points of $A$, which I am finding trouble with. Any help would be appreciated. Thanks

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  • $\begingroup$ Is this a correct drawing of $A$? $\endgroup$
    – DHMO
    Apr 19, 2017 at 17:07

1 Answer 1

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Let $\overline{B}_r(x)$ denote the closed Ball with radius $r$ and center $x$. Then it is pretty clear that for the closure of $A$ denoted by $\overline{A}$ the following holds true \begin{align*} \overline{A} = \overline{\bigcup_{n\in\mathbb{N}} B_{\frac{1}{n}}\!\!\left(\frac{1}{n},n\right)} \supseteq \bigcup_{n\in\mathbb{N}} \overline{B}_{\frac{1}{n}}\!\!\left(\frac{1}{n},n\right) =: C \end{align*} Now it's left to show that the set $C$ is already closed. Therefore take an arbitrary convergent sequence $(x_n)_{n\in\mathbb{N}}$ such that $x_n \in C$ for every $n\in\mathbb{N}$. Let $x$ denote the limit of the sequence. Then we want to show that $x$ must already be in $C$:

Because $x_n$ tend to $x$ there is a $n_0$ such that \begin{align*} d(x,x_n) < \frac{1}{100}\quad \text{for all}\quad n\geq n_0 \end{align*} Now we can conclude that the set $\{x_k : k\geq n_0\}$ can at most intersect two balls "of" $C$. We denote these two balls as $B_1$ and $B_2$. Since both $B_1$ and $B_2$ are closed the limit $x$ must already be in one of them and therefore in $C$.

This shows that $C$ is closed. We already stated $C \subseteq \overline{A}$. Combined with the obvious fact $A\subseteq C$ we receive \begin{align*} \overline{A} = C \end{align*}

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