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

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

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dextercioby
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(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
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.