#(=>) :#
Let M be a compact subset of a finite dimensional normed space X and let #x in barM#
#therefore EE# a sequence #(x_n)# in M, converging to x.
#therefore# every subsequence #(x_(nk))# of #(x_n)# also converges to x.
#therefore x in M#
#therefore bar M = M =># M is closed.
Suppose M is not bounded. (Wanting : M not compact. ie. the contrapositive proof).
Choose any #x in X#
#therefore EE# a sequence #(x_n)# in M such that #||x_n-x||>n AA n#
#therefore# this sequence has no convergent subsequence.
#therefore# M is not compact.
This is a contradiction to the given premise that M is compact.
Hence it follows that M is bounded.
#(lArr )# :
Assume M is closed and bounded.
#therefore EE k in RR# such that #||x|| <= k AA x in M#
Let #{e_1, e_2, ..., e_n} # be a basis for X.
Let #(y_n)# be any sequence in M.
Hence, assuming dim(X) = n, each #y_m# can be expressed uniquely in the form
#y_m=alpha_(1m)e_1 + alpha_(2m)e_2 +... + alpha_(nm)e_m#
Therefore by a lemma of normed spaces, #EE c in RR^+# such that #k>=||y_m||>=c sum_(j=1)^n |alpha_(jm)| AA j and AA m=1,2,3,....,n#
Therefore the sequence #(alpha_(jm))# in #RR# is bounded in M since #(alpha_(jm))<=k/c#
#therefore# by the Bolzano-Weierstrass Theorem, there exists a convergent subsequence converging to #alpha_1# say.
Let #(y_(1m))# be the corresponding subsequence of #(y_m)#
Now consider the corresponding subsequence of #(alpha_(2m))#.
Since it is bounded, it has a convergent subsequence converging to #alpha_2# say.
Let #(y_(2m))# be the corresponding subsequence of #(y_(1m))#
Continuing in this way, we eventually get a subsequence #(y_(nm))# of #(y_m)# such that
#y_(nm)=gamma_(1m)e_1+gamma_(2m)e_2+ ..... + gamma_(nm)e_n#, with the property that #(gamma_(jm))_(m=1,2,3,....)# converges to #gamma_j AA j=1,2,3, .... , n#
Let #y=alpha_1e_1+alpha_2e_2+.......+alpha_n e_n#
#therefore ||y_(nm)-y||=||(gamma_(1m)-alpha_1)e_1+(gamma_(2m)-alpha_2)e_2+......+(gamma_(nm)-alpha_n)e_n||#
#<= sum_(j=1)^n |gamma_(jm)-alpha_j|*||e_j||#
#-> 0# as #m -> oo#
#therefore (y_(nm)) -> y#
But since M is closed, #=> y in M#
#therefore# M is compact.
