Exercise:
Let V be a finite dimensional vector space over K. Then V has a basis with finitely many elements. Moreover every basis of V is finite and has the same number of elements. Theorem is true even if V is not finite dimensional so called glqq Hamel basis theoremgrqq

Solution:
Proof. For the proof of the Theorem it is helpful to first show some Lemmas which can then be used. bf Lemma E: Let w_..w_lin V and ase that w_jnotin Spw_..w_j- forall leq jleq l. Then the list w_...w_l is linearly indepent. Proof of Lemma E. If w_..w_l were linearly depent then by Lemma A exists leq jleq l s.t. w_jin Spw_...w_j- contradiction lightning bf Lemma F: Suppose that Spv_..v_nV. Then exists a subset of v_...v_n which is a basis for V. Proof of Lemma F. The proof is algorithmic and is based on n steps. At each step we'll consider another vector from the list v_...v_n and decide whether or not to drop it from the list. At step #j we consider v_j. bf step : If v_ we drop v_ from the list. If v_neq we keep it. bf step : If v_jin Spv_...v_j- we drop v_j. Otherwise we keep it. After performing n steps as above we obtain a new possibly shorter list: v_i_v_i_...v_i_m where leq i_ i_ ... i_mleq n. We claim that Spv_i_...v_i_mV. Indeed in any of the steps leq jleq n we dropped the vector v_j only if v_jin Spv_...v_j-. Longrightarrow Spv_i_...v_i_mSpv_..v_nV. We also claim that v_i_...v_i_m are linearly indepent. Indeed forall leq kleq m we have v_i_knotin Spv_v_...v_i_k- Longrightarrow v_i_k notin Spv_i_...v_i_k-. By Lemma E v_i_...v_i_m is linearly indepent. Together with the fact that Spv_i_...v_i_mV we get that v_i_...v_i_m is a basis for V. We can now prove the theorem. Proof. The theorem states three things: itemize item V has a basis with finitely many elements. item Every basis of V has finitely many elements. item Every two bases mathcalA and mathcalB are finite and moreover |mathcalA||mathcalB|. itemize bf First statement: Follows from Lemma F. Take any finite set S s.t. SpSV such an S exists because by assption V is finite dimensional. By Lemma F a subset S' subseteq S is a basis for V. bf Second statement: Every basis mathcalC of V is finite. Suppose by contradiction that mathcalC is a basis of V but mathcalC is not a finite set. Let S be a finite basis of V such S exists by asption say Sv_...v_n. Choose n+ vectors u_...u_n+in mathcalC. They are linearly indepent. By Lemma D n+leq n contradiction lightning This shows that every basis mathcalC of V is finite. bf Third statement: Indeed by the second statement |mathcalA||mathcalB| infty. Now SpBV and mathcalA is linearly indepent so by Lemma D we get that |mathcalA|leq|mathcalB|. But also SpAV and mathcalB is linearly indepent so by Lemma D again we get that |mathcalB|leq|mathcalA|. This shows SpBV and mathcalA is linearly indepent so by Lemma D we get that |mathcalA||mathcalB|.