Lemma: If

\\\\phi :N->N

is a strictly increasing function, then for\ all

ninN

we have

\\\\phi (n)>=n

.\ Proof of lemma: Let

\\\\phi :N->N

be a strictly increasing function. We\ prove the claim by induction on

n

.\ For the base case,

n=1

, it suffices to note that\

>=, since \\\\phi (1)inN.

\ For the induction step, assume that for a natural number

ninN

we\ have

\\\\phi (n)>=n

. We must prove the corresponding property for

n+1

.\ Since

\\\\phi

is strictly increasing, we have\

>

\ which implies

\\\\phi (n+1)>=\\\\phi (n)+1

, since the strict inequality above\ concerns natural numbers. To conclude the proof, we estimate\

\\\\phi (n+1)>=\\\\phi (n)+1>=

\ using\ in the last step.\ Theorem: If

(x_(n))_(ninN)

is a sequence of real numbers which converges\ to

\\\\alpha inR

, then any subsequence of it also converges to

\\\\alpha

.\ Proof of theorem: Let

(x_(n))_(ninN)

be a sequence such that

\\\\lim_(n->\\\\infty )x_(n)=\\\\alpha

,\ and let

(x_(\\\\phi (n)))_(ninN)

be a subsequence of it. We must prove\

\\\\lim_(n->\\\\infty ),=\\\\alpha

\ To prove this from the definition of limit, let

\\\\epsi >0

. From the\ assumption

\\\\lim_(n->\\\\infty )x_(n)=\\\\alpha

we get that there exists an

NinN

\ such that for all\ we have\ The same

N

will work for our present goal. So let

n>=N

. Note that\ by the lemma we then get

\\\\phi (n)>=n>=N

. In particular from the\ above, we obtain

|x_(\\\\phi (n))-\\\\alpha |

. This finishes the proof, directly by\ the definition of a limit.

Lemma: If :NN is a strictly increasing function, then for all nN we have (n)n. Proof of lemma: Let :NN be a strictly increasing function. We prove the claim by induction on n. For the base case, n=1, it suffices to note that since(1)N. For the induction step, assume that for a natural number nN we have (n)n. We must prove the corresponding property for n+1. Since is strictly increasing, we have which implies (n+1)(n)+1, since the strict inequality above concerns natural numbers. To conclude the proof, we estimate (n+1)(n)+1 using in the last step. Theorem: If (xn)nN is a sequence of real numbers which converges to R, then any subsequence of it also converges to . Proof of theorem: Let (xn)nN be a sequence such that limnxn=, and let (x(n))nN be a subsequence of it. We must prove limn To prove this from the definition of limit, let >0. From the assumption limnxn= we get that there exists an NN such that for all we have The same N will work for our present goal. So let nN. Note that by the lemma we then get (n)nN. In particular from the above, we obtain x(n)<. this finishes the proof directly by definition of a limit>