I am taking an introductory topology class, and we recently defined the notion of compactness. Earlier in the chapter, the Koch snowflake is described, and I am wondering: is the Koch snowflake a compact set? Intuitively I think the answer is yes: it is an infinite union of closed sets (not necessarily closed) and its limiting area is bounded, so it is a bounded and closed set of $\mathbb{R}^{2}$. Am I just waiving my hands, or is this a solid argument?
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A.E
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5An infinite union of closed sets is not necessarily closed. For example, $\bigcup_{n=1}^{\infty}[\frac{1}{n}, 1] = (0, 1]$. – Michael Albanese Oct 16 '13 at 23:10
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2Don’t waive your hands: you might need them! :-) – Brian M. Scott Oct 16 '13 at 23:12
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Thanks @MichaelAlbanese, that was a slip on my part. – A.E Oct 16 '13 at 23:23
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The union of infinitely many closed sets is not necessarily closed. However, the Koch snowflake is compact: it’s the range of a continuous function defined on the compact set $[0,1]$, and continuous functions preserve compactness.
Brian M. Scott
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What is the continuous function between the two spaces? I am worried that just "saying" there is a continuous function between them is not enough. – A.E Oct 16 '13 at 23:22
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