Re: Can we say (0, 1) is compact in (0, 1)?
- From: "W. Dale Hall" <wd_hall@xxxxxxxxxxx>
- Date: Wed, 30 Nov 2005 07:58:46 GMT
comtech wrote:
Hi all,
Suppose my metric space M is (0, 1),
and set A is (0, 1) in M,
can we say that A is compact (in M)?
My reasoning is that:
A is both closed and open in M, since A=M, let's deem it as closed; A is bounded too, So A is bounded and closed => A is compact, by the Heine-Borel theorem...
Am I correct? I am really not sure...
We can say anything we like, but (0,1) is not compact with the standard topology. Compactness is defined as follows:
X is compact if, for every family
U = { U_j | j in J }where U_j's are all open, and J is an arbitrary indexing set
such that X = union(U_j | j in J)
there is a finite subset Jo of J, for which X = union(U_j | j in Jo}.
This is typically related by saying "every open cover has a finite subcover", the term "cover" referring to a family of sets having union equal to the entire space.
That (0,1) is not compact can be seen by considering the open cover
U = {(1/n, 1-1/n) | n = 3,4,5,... }.Note that the union of all the intervals (1/n , 1-1/n) is (0,1), but no finite subset of this family covers the entire interval (0,1).
Your version of compact is in reality the condition stated in the Heine-Borel theorem, with the exception that you omitted the context, namely that the set be considered as a subset of R^n. Without that context, the characterization does not capture the essential notion of compactness.
Dale. .
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