Is the close interval A=[0,1] is compact?

[rohan302]

Is the close interval A=[0,1] is compact?

 

[klackity]

Yes.

This is a result of the Heine-Borel theorem: http://en.wikipedia.org/wiki/Heine–Borel_theorem

 

[dextercioby]

(0,1) is not closed, but it's bounded. So taking its closure in the interval metric one gets the closed interval hence the compactness property.

 

[HallsofIvy]

Assuming you are talking about the "usual topology" on the real numbers (the metric topology defined by the metric d(x,y)= |x- y|) then, yes, that set is both closed and bounded and the Heine-Borel theorem applies, so it is compact.

But it is necessary to specify the topology, not just the set. While the topology I cited above is the "usual" topology, we could also give the set of all real numbers the "discrete" topology which is the metric topology defined by "d(x, x)= 0 but if $x\ne y$ d(x,y)= 1". Then it is easy to show that every set is closed and every set is bounded but the only compact sets are the finite sets. In that topology, [0, 1] is both closed and bounded but is not compact. Obviously, the "Heine-Borel theorem" does not apply in that topology.