Can you please help me with the question:
How to proof that a bounded, open interval is not compact.
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$\begingroup$ i think you need to write your definition of compactness . $\endgroup$– bytrzFeb 2, 2015 at 23:19
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4 Answers
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That depends on your definition of compactness.
For $(a,b)$, either consider the open sets $\{(a,b-1/n):n \in \Bbb N\}$ or the sequence $x_n = b - 1/n$ (with $n \in \Bbb N$).
answered Feb 2, 2015 at 23:12
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Which space? Do you mean $\mathbb{R}^n$? Recall the definition of compactness in $\mathbb{R}^n$. $$ A \subset \mathbb{R}^n \text{is compact, iff} A \text{is closed and bounded} $$ You can show that there is a sequence, such that the limit point is not contained in your open intervall.
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$\begingroup$ Surely that's not the textbook definition, since that claim is false in general. $\endgroup$– user4894Feb 2, 2015 at 23:19
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$\begingroup$ yeah, in genereal. thats why i asked : do you mean $\mathbb{R}^n$. of course i mean $\mathbb{R}^n$ as a metric space. $\endgroup$– aGerFeb 2, 2015 at 23:25
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If $\,I=\,(\,a,b\,)\,$ is such an interval, the sequence of intervals $\,(a,a+\frac1{2^n}]$ is a decreasing sequence of closed intervals of the topological space $I$, and the intersection of all these closed intervals is empty.
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Hint: Consider the cover $A=\bigcup_{n\in \mathbb{N}}\{(\frac{1}{n},1)\}$ for the interval $(0,1) \subset \mathbb{R}$.
answered Feb 2, 2015 at 23:12