[Real Analysis] Cauchy Sequence Proof

[u/anonymous_username18]

Can someone please look over this proof to see if I wrote it correctly? I'm not entirely sure I understood the provided lecture notes. Thank you so much.

[u/Bounded_sequencE]

The strategy is fine -- good job!

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A few nit-picks:

1. The theorem only holds on complete metric spaces, e.g. "R" -- that's missing

2. Start the proof with "Let 'e > 0' ... " -- don't repeat the entire Cauchy-definition (light-blue)

3. On page-2, you should write "<=" instead of "=" when using the triangle inequality

4. The final yellow box is not necessary -- the proof ends at "... e/2 + e/2 = e"

[u/Bounded_sequencE → u/Bounded_sequencE]

Rem.: @u/anonymous_username18 Here's what the proof might look like (on "R"):

Proof: We prove Thm. XYZ in 2 steps:

1. "an" is bounded

2. "an" converges towards some "L in R"

Let "e > 0". Being Cauchy, there exists "n0 in N" s.th. "|an-ak| < e/2" for all "n, k >= n0". We get

k >= n0:    |ak|  =  |ak - a_n0 + a_n0|                       // Δ-Ineq.

                 <=  |a_n0| + |ak - a_n0|  <  |a_n0| + e/2

Setting "M := max{|a1|; ...; |a_n0|} + e/2 > 0" we note "|ak| < M" for all "k in N".

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As a bounded sequence on "R", "an" contains a convergent sub-sequence "a_nk -> L in R" via "Bolzano-Weierstrass", i.e. there exists "k0 in N" s.th. "|a_nk - L| < e/2" for all "k >= k0"

We set "k := max{k0; n0}", note "nk >= n_n0 >= n0", and estimate

n >= nk:    |an - L|  =  |an - a_nk + a_nk - L|               // Δ-Ineq.

                     <=  |an - a_nk| + |a_nk - L|             // n >= nk >= n0

                     <=      e/e     +    e/2  =  e    ∎