(New page: 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 <math>\mathbf{R}</math>, so say the metric mapped ...)
 
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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 <math>\mathbf{R}</math>, so say the metric mapped to <math>\mathbf{Q}</math>.  Then certainly it could be the case that there was even a point in <math>X</math>, let alone <math>K</math>, where <math>d(x,k) \inf bla</math>.  [[User:Dimberti|Dimberti]] 15:55, 21 September 2008 (UTC)
 
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 <math>\mathbf{R}</math>, so say the metric mapped to <math>\mathbf{Q}</math>.  Then certainly it could be the case that there was even a point in <math>X</math>, let alone <math>K</math>, where <math>d(x,k) \inf bla</math>.  [[User:Dimberti|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.  [[User:Dimberti|Dimberti]] 15:56, 21 September 2008 (UTC)

Revision as of 10:56, 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)

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang