Dtjytyk

Are all namekians brothers?

What is this high flying aircraft over Pennsylvania?

Has the laser at Magurele, Romania reached a tenth of the Sun's power?

Calculate Pi using Monte Carlo

New Order #2: Turn My Way

Should I warn a new PhD Student?

What should be the ideal length of sentences in a blog post for ease of reading?

Reason why a kingside attack is not justified

Could a welfare state co-exist with mega corporations?

Output visual diagram of picture

How to test the sharpness of a knife?

What is the tangent at a sharp point on a curve?

Hashing password to increase entropy

What do the positive and negative (+/-) transmit and receive pins mean on Ethernet cables?

What is it called when someone votes for an option that's not their first choice?

Connection Between Knot Theory and Number Theory

Did I make a mistake by ccing email to boss to others?

Is divisi notation needed for brass or woodwind in an orchestra?

Can you take a "free object interaction" while incapacitated?

Friend wants my recommendation but I don't want to give it to him

Would a primitive species be able to learn English from reading books alone?

Does capillary rise violate hydrostatic paradox?

Why didn’t Eve recognize the little cockroach as a living organism?

Is there a POSIX way to shutdown a UNIX machine?

Is this proof of a theorem on subsequence convergence correct?

Counterexamples to proofs of correct statementsFundamental theorem of algebra: a proof for undergrads?Alternate proof for a theorem on ordered fieldsProving convergence by convergence of 3 subsequencesIdentifying a property of contiuous of functions and its proofIs my own proof of the Bolzano-Weierstrass Theorem correct?Proof of 'sandwich theorem' for sequencesProving the set of all limit points of a set is closed - alternative proof.Theorem Explanation: Sequences and ConvergenceIs my proof of “Theorem 1.2” correct?

0

$begingroup$

I'm taking my first course on analysis. I came across this theorem and I wonder whether this proof that I have written is correct. In case it is incorrect, please point out the mistakes I have made.

Here is the written proof:

real-analysis analysis proof-writing

share|cite|improve this question

edited Mar 12 at 14:44

Arthur

118k7118202

asked Mar 12 at 14:43

Shivam BauraiShivam Baurai

11

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is k in the second to last line (for all n greater than k...)? Also, can you explain the definition of S'? The notation is a bit confusing.
  $endgroup$
  – FelixMP
  Mar 12 at 19:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @FélixMorenoPeñarrubia k is a natural number. Perhaps that and S' would be better understood if we follow the definition of a subsequence. I've used exactly the definition but in terms of a set because, personally, I find the concept of subsequences much easier to grasp in those terms.
  $endgroup$
  – Shivam Baurai
  Mar 13 at 12:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I meant to ask what does k represent. You use a variable named "k" in the previous line when you state the definition of convergence of a subsequence, but I see no relationship with the "for all n greater than k" statement in the second-to-last line. I can't understand either what roles do the sets S and S' play in that step. I would recommend to rewrite and clarify the second part of the proof, as it is currently not understandable.
  $endgroup$
  – FelixMP
  Mar 13 at 15:36
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, let's see, X', the subsequence of X, is also a sequence and will thus have the same definition of convergence as sequences. Now, in that definition, we talk about the terms of (x¬n) after a certain m (we say "for all n ≥ m”). Similarly for subsequence X', the definition would hold for all nk ≥ m, which is similar to saying k ≥ m, because we know nk ≥ k for all k.
  $endgroup$
  – Shivam Baurai
  Mar 13 at 17:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As for the set notation, as I said, it helps ME, especially in being convinced about making the ‘transition’ from xnk to xn(which we can do because as S' is a subset of S, so xnk belongs to X and is also of the form xn), that we make in the second last line of the proof. As for the problem you have with n ≥ k, this comes because I want to say n ≥ m, and since k ≥ m, n ≥ k works just fine.
  $endgroup$
  – Shivam Baurai
  Mar 13 at 17:38
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

I'm taking my first course on analysis. I came across this theorem and I wonder whether this proof that I have written is correct. In case it is incorrect, please point out the mistakes I have made.

Here is the written proof:

real-analysis analysis proof-writing

share|cite|improve this question

edited Mar 12 at 14:44

Arthur

118k7118202

asked Mar 12 at 14:43

Shivam BauraiShivam Baurai

11

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is k in the second to last line (for all n greater than k...)? Also, can you explain the definition of S'? The notation is a bit confusing.
  $endgroup$
  – FelixMP
  Mar 12 at 19:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @FélixMorenoPeñarrubia k is a natural number. Perhaps that and S' would be better understood if we follow the definition of a subsequence. I've used exactly the definition but in terms of a set because, personally, I find the concept of subsequences much easier to grasp in those terms.
  $endgroup$
  – Shivam Baurai
  Mar 13 at 12:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I meant to ask what does k represent. You use a variable named "k" in the previous line when you state the definition of convergence of a subsequence, but I see no relationship with the "for all n greater than k" statement in the second-to-last line. I can't understand either what roles do the sets S and S' play in that step. I would recommend to rewrite and clarify the second part of the proof, as it is currently not understandable.
  $endgroup$
  – FelixMP
  Mar 13 at 15:36
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, let's see, X', the subsequence of X, is also a sequence and will thus have the same definition of convergence as sequences. Now, in that definition, we talk about the terms of (x¬n) after a certain m (we say "for all n ≥ m”). Similarly for subsequence X', the definition would hold for all nk ≥ m, which is similar to saying k ≥ m, because we know nk ≥ k for all k.
  $endgroup$
  – Shivam Baurai
  Mar 13 at 17:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As for the set notation, as I said, it helps ME, especially in being convinced about making the ‘transition’ from xnk to xn(which we can do because as S' is a subset of S, so xnk belongs to X and is also of the form xn), that we make in the second last line of the proof. As for the problem you have with n ≥ k, this comes because I want to say n ≥ m, and since k ≥ m, n ≥ k works just fine.
  $endgroup$
  – Shivam Baurai
  Mar 13 at 17:38
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I'm taking my first course on analysis. I came across this theorem and I wonder whether this proof that I have written is correct. In case it is incorrect, please point out the mistakes I have made.

Here is the written proof:

real-analysis analysis proof-writing

share|cite|improve this question

edited Mar 12 at 14:44

Arthur

118k7118202

asked Mar 12 at 14:43

Shivam BauraiShivam Baurai

11

$endgroup$

share|cite|improve this question

edited Mar 12 at 14:44

Arthur

118k7118202

asked Mar 12 at 14:43

Shivam BauraiShivam Baurai

11

share|cite|improve this question

edited Mar 12 at 14:44

Arthur

118k7118202

asked Mar 12 at 14:43

Shivam BauraiShivam Baurai

11

edited Mar 12 at 14:44

Arthur

118k7118202

edited Mar 12 at 14:44

Arthur

118k7118202

asked Mar 12 at 14:43

Shivam BauraiShivam Baurai

11

asked Mar 12 at 14:43

Shivam BauraiShivam Baurai

11