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i) Make the argument that with the infimum you can construct a sequence that approaches it. | i) Make the argument that with the infimum you can construct a sequence that approaches it. | ||
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ii) I'm sure there's a theorem somewhere that states that point of a limit implies limit point. | ii) I'm sure there's a theorem somewhere that states that point of a limit implies limit point. | ||
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iii) K is compact so it's closed. | iii) K is compact so it's closed. | ||
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iv) Since it's closed, the limit point is in K. | iv) Since it's closed, the limit point is in K. | ||
The problem I have with this is that I'm kind of meta-reading the question, so it makes the next part trivial. [[User:Dimberti|Dimberti]] 16:06, 21 September 2008 (UTC) | The problem I have with this is that I'm kind of meta-reading the question, so it makes the next part trivial. [[User:Dimberti|Dimberti]] 16:06, 21 September 2008 (UTC) | ||
Latest revision as of 11:07, 21 September 2008
Got 1 and 2 in class. I was thinking about number 3, and got into this line of thinking. We don't know that the metric is surjective to $ \mathbf{R} $, so say the metric mapped to $ \mathbf{Q} $. Then certainly it could be the case that there was even a point in $ X $, let alone $ K $, where $ d(x,k) \inf bla $. Dimberti 15:55, 21 September 2008 (UTC)
- Ah, but then, just like in the discrete metric, the only compact sets would be finite ones. Nevermind. Dimberti 15:56, 21 September 2008 (UTC)
O.k., here was the line of thought I was going on, but am dubious about its success.
i) Make the argument that with the infimum you can construct a sequence that approaches it.
ii) I'm sure there's a theorem somewhere that states that point of a limit implies limit point.
iii) K is compact so it's closed.
iv) Since it's closed, the limit point is in K.
The problem I have with this is that I'm kind of meta-reading the question, so it makes the next part trivial. Dimberti 16:06, 21 September 2008 (UTC)
