Abstract
Therefore some subsequence has every term in K j . Since K j is compact, it follows that the subsequence has a subsequence converging to a limit in K j , which must then belong to the union K. But this subsequence is itself a subsequence of the sequence that we started with. The latter is therefore seen to have a subsequence converging to a limit in K.
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© 2011 Springer-Verlag London Limited
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Shirali, S., Vasudeva, H.L. (2011). Solutions. In: Multivariable Analysis. Springer, London. https://doi.org/10.1007/978-0-85729-192-9_9
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DOI: https://doi.org/10.1007/978-0-85729-192-9_9
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