Proof. Suppose not. Then for all natural numbers we can find some such that . The sequence is bounded, so by the Bolzano-Weierstrass theorem it has a convergent sub sequence, say . As is closed converges to a value in . By the continuity of we should have that converges, but by construction it diverges. This contradiction finishes the proof.
|Date of creation||2013-03-22 14:29:18|
|Last modified on||2013-03-22 14:29:18|
|Last modified by||classicleft (5752)|