I read book of Munkres in Separation theorem in Plane , he said : " Let $C$ be a compact subspace of $S^{2}$ ". Then why $S^{2}-C$ is locally connected ? It's $61.1$ theorem . Sorry for my bad English
Why $S^{2} - C$ is locally connected
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general-topology
1 Answers
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Because $S^2$ is locally connected, and an open subspace of a locally connected space is locally connected. $C$ is compact, so closed, and so $S^2 - C$ is an open subspace of $S^2$. Hence the claim.
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0So why $U$ is a component of $S^{2} - C$ then $U - b$ is connected with $b$ is a point in $S^{2} - C$ . I can't understand soluntion in Munkres's book . Can you explain it for me ? – 2017-01-14