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Message |
| G.C. |
 Posted: Fri Dec 30, 2005 7:03 am |
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Guest
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Let A and B be two compact sets in R^n.
Then I think that
A - B={a-b : a in A, b in B}
is also compact in R^n.
As a subset of R^n, Heine-Borel suggest it suffices to prove that
A-B is both closed and bounded.
Or, by definition, given any collection of open sets covering A-B, it's
possible to costruct a finite sub-covering.
Can you help me in proving that A - B is compact, please?
Thanks,
G.C. |
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| Jannick Asmus |
| Posted: Fri Dec 30, 2005 7:13 am |
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Guest
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On 30.12.2005 13:03, G.C. wrote:
[quote:ac59962bb6]Let A and B be two compact sets in R^n.
Then I think that
A - B={a-b : a in A, b in B}
is also compact in R^n.
As a subset of R^n, Heine-Borel suggest it suffices to prove that
A-B is both closed and bounded.
Or, by definition, given any collection of open sets covering A-B, it's
possible to costruct a finite sub-covering.
Can you help me in proving that A - B is compact, please?
Thanks,
G.C.
[/quote:ac59962bb6]
You could use the fact that the image of a compact set under a
continuous mapping is compact again (in Hausdorff spaces).
J. |
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| Guest |
| Posted: Fri Dec 30, 2005 8:18 am |
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To use Asmus' idea, which works even without Hausdorff assumptions, you
will have to interpret addition (or subtraction) as a continuous
function from X^2 to X, where X is R^n. You will also have to use the
fact that cartesian products of compact spaces are again compact.
If this sounds unfamiliar, for example if you don't know what a
cartesian product is, we can use the characterization of compact sets
in R^n as those that are closed and bounded. Closure in a metric space
is the same as sequential closure. You ought to be able to see A- B is
bounded fairly readily. You need to verify it is closed, so you verify
that any sequence in A-B that converges has a limit in A-B. Basically:
suppose (a_n - b_n) converges. You need to show it converges to
something in A - B. The problem here is that a_n need not converge in A
and likewise b_n need not converge in B. Now heres the HINT: every
sequence in a compact space has a convergent subsequence. However, it
will be tricky to apply this hint because you need to find a
subsequence so that both a_(n_k) and b_(n_k) are both convergent
subsequences. I've said enough, probably too much. Good luck fleshing
out the details. |
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| Guest |
| Posted: Sat Dec 31, 2005 1:52 pm |
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| Is the difference [0,4] - [2,3] compact? |
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| Ed Hook |
| Posted: Sat Dec 31, 2005 2:00 pm |
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barr@barrs.org wrote:
[quote:11df35310f]Is the difference [0,4] - [2,3] compact?
[/quote:11df35310f]
[0,4] - [2,3] = [-3,2] (using the notation in the original post), so
....
"Yes" |
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