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.

 

[blue_raver22]

Yes, the closed interval A=[0,1] is compact. This means that it is both closed and bounded, and therefore contains all of its limit points. In other words, any sequence of points within the interval will have a limit point that is also within the interval. This property is important in many areas of mathematics and physics, as it allows for the use of certain theorems and techniques that rely on compactness.