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Consider the interval $ S_k = (1 − 1/k, 2 + 1/k] $ and I want to find $ \cup^\infty_{k=1} S_k $ with proof

How do I go about this?

What I was thinking... as $ k \rightarrow \infty $ $ 1/k \rightarrow 0 $ so the biggest interval will be (0, 3]. And every interval past $ k = 1 $ will be smaller than this interval hence the union must be equal to (0, 3] since $inf S_k = 0 $ and $ sup S_k = 3 $ and every point within these smaller intervals is necessarily contained in this interval?

Don't know if that's technically correct or not, and if it is how to go about putting that down formally. I don't have limits yet so I cant really state 1/k approaches 0?

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Hint

$$\bigcup_{k=1}^\infty S_k=\{x\mid \exists k\in\mathbb N^*: x\in S_k\}$$

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  • $\begingroup$ Since $S_k\subset S_0$ for all $k$, the answer looks obvious. $\endgroup$ – Surb Feb 4 '16 at 12:29
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It is enough to state:

  • $(0,3]=S_1\subseteq\bigcup_{k=1}^{\infty}S_k$.
  • $S_k\subseteq(0,3]$ for each $k\in\{1,2,\dots\}$, hence $\bigcup_{k=1}^{\infty}S_k\subseteq(0,3]$.

This together allows the conclusion:$$(0,3]=\bigcup_{k=1}^{\infty}S_k$$

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  • $\begingroup$ Note that the$S_k$ are right-closed, so it's $(0,3]$ actually $\endgroup$ – Hagen von Eitzen Feb 4 '16 at 12:43
  • $\begingroup$ @HagenvonEitzen Thank you, I repaired. $\endgroup$ – drhab Feb 4 '16 at 14:17

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