Vector Extreme Value Theorem

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(Difference between revisions)
(Vector Extreme Value Theorem)
(Vector Extreme Value Theorem)
 
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  <m>\\right F(A)</m> is compact [i.e. F is bounded and attains its 'extreme' values]
  <m>\\right F(A)</m> is compact [i.e. F is bounded and attains its 'extreme' values]
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And apparently, there's no analogue to the [[Intermediate Value Theorem]] in Vector Calculus.
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And apparently, there's no analogue to the [[Intermediate Value Theorem]] in Vector Calculus. (CZ: As you have probably seen from the notes, that's as relative as the word ''intermediate''. The topological Extreme Value Theorem says that a continuous function maps a compact set to a compact set, and the Intermediate Value Theorem says that a continuous function maps a connected set to a connected set. Put them together and you get that a continuos function from <m>\\bbR</m> to <m>\\bbR</m> sends a closed interval to a closed interval.)
And then I have these scribblings:
And then I have these scribblings:

Current revision as of 00:17, 12 February 2006

[edit] Vector Extreme Value Theorem

<m>F: \\bbR^n \\right \\bbR^m</m>
<m>F</m> is continuous, <m>A \\subset \\bbR^n</m> compact subset
<m>\\right F(A)</m> is compact [i.e. F is bounded and attains its 'extreme' values]

And apparently, there's no analogue to the Intermediate Value Theorem in Vector Calculus. (CZ: As you have probably seen from the notes, that's as relative as the word intermediate. The topological Extreme Value Theorem says that a continuous function maps a compact set to a compact set, and the Intermediate Value Theorem says that a continuous function maps a connected set to a connected set. Put them together and you get that a continuos function from <m>\\bbR</m> to <m>\\bbR</m> sends a closed interval to a closed interval.)

And then I have these scribblings:

If <m> F(A)\\subset \\bigcup{i \\in I}{}{U_i} </m> is an open covering of <m>F(A)</m>

then <m> A \\subset \\bigcup{i \\in I}{}{F^{-1}(U_i)} </m>

and since <m>F</m> is continuous, the inverse images are open sets, so we have an open covering of <m>A</m>. Since <m>A</m> is compact, there is a finite subcovering

<m> A \\subset \\bigcup{k=1}{n}{F^{-1}(U_i_k)} </m>

and therefore

<m>F(A) \\subset \\bigcup{k=1}{n}{(U_i_k)}</m> is a finite subcovering of the initial open covering.

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