Re: Can we say (0, 1) is compact in (0, 1)?



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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