Fdtxjr

Bete Noir -- no dairy

How to answer "Have you ever been terminated?"

The logistics of corpse disposal

Why did the IBM 650 use bi-quinary?

Dating a Former Employee

Why didn't this character "real die" when they blew their stack out in Altered Carbon?

Is there a (better) way to access $wpdb results?

Apollo command module space walk?

Identifying polygons that intersect with another layer using QGIS?

51k Euros annually for a family of 4 in Berlin: Is it enough?

How come Sam didn't become Lord of Horn Hill?

Why is "Consequences inflicted." not a sentence?

Using et al. for a last / senior author rather than for a first author

How to call a function with default parameter through a pointer to function that is the return of another function?

Single word antonym of "flightless"

String `!23` is replaced with `docker` in command line

Can a USB port passively 'listen only'?

Sci-Fi book where patients in a coma ward all live in a subconscious world linked together

Why aren't air breathing engines used as small first stages

Why do we bend a book to keep it straight?

Do I really need recursive chmod to restrict access to a folder?

Is the Standard Deduction better than Itemized when both are the same amount?

Overriding an object in memory with placement new

How to react to hostile behavior from a senior developer?

Proving any finite non-empty set of real numbers has a max and a min.

1

$begingroup$

My proof goes as follows:

Let S be a finite non-empty set of real numbers with exactly n elements. Assume by contradiction that S does not a max value.

That is, for every element in S there exists another element in S that is strictly greater than it.

Note that these two must be distinct since the latter will be strictly greater than the former.

Now, choose any element in S, and denote it by x_1. By our assumption, there is a distinct element in S strictly greater than x_1, which we denote by x_2.

Continuing in this fashion, for x_n-1 in S, there is x_n in S strictly greater than x_n-1. However, by our assumption there will also exist x_n+1 in S strictly greater than X_n.

Since x_n+1 cannot be x_n and since S has exactly n elements, then x_n+1 must be one of the following: x_1 , x_2, ..... x_n-1. However, these are all strictly less than x_n. So, this is impossible. That is, we have a contradiction in any case.

The case for the min value of S follows similarly.

My questions are

1. Is this proof correct?




2. How do I formalize it? Assuming this isn't rigorous enough.

real-analysis

share|cite|improve this question

edited Mar 26 at 17:02

J. W. Tanner

4,9371420

asked Mar 26 at 16:26

Gabriel OlivoGabriel Olivo

63

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A non-empty set of what?
  $endgroup$
  – José Carlos Santos
  Mar 26 at 16:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Elements from what kind of set? Totally ordered? Poset? Lattice?
  $endgroup$
  – Wuestenfux
  Mar 26 at 16:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should add the word $finite$ to your title.
  $endgroup$
  – John Douma
  Mar 26 at 16:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might need to prove that if $A$ is a finite set with $n$ elements then it can not be a finite set with $m;m ne n$ elements. Or that a set can't be both finite and infinitely countable. Or simply a set must have one specific cardinality and not two. Also twice you seem to claim elements are distinct because they are strictly greater. It's enough to say they are distinct because elements of sets are distinct by definition.
  $endgroup$
  – fleablood
  Mar 26 at 17:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can sort the numbers, so it is obvious that the set must have a minimum and a maximum (which coincide in the case the numbers are all equal)
  $endgroup$
  – Peter
  Mar 26 at 17:53
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

My proof goes as follows:

Let S be a finite non-empty set of real numbers with exactly n elements. Assume by contradiction that S does not a max value.

That is, for every element in S there exists another element in S that is strictly greater than it.

Note that these two must be distinct since the latter will be strictly greater than the former.

Now, choose any element in S, and denote it by x_1. By our assumption, there is a distinct element in S strictly greater than x_1, which we denote by x_2.

Continuing in this fashion, for x_n-1 in S, there is x_n in S strictly greater than x_n-1. However, by our assumption there will also exist x_n+1 in S strictly greater than X_n.

Since x_n+1 cannot be x_n and since S has exactly n elements, then x_n+1 must be one of the following: x_1 , x_2, ..... x_n-1. However, these are all strictly less than x_n. So, this is impossible. That is, we have a contradiction in any case.

The case for the min value of S follows similarly.

My questions are

1. Is this proof correct?




2. How do I formalize it? Assuming this isn't rigorous enough.

real-analysis

share|cite|improve this question

edited Mar 26 at 17:02

J. W. Tanner

4,9371420

asked Mar 26 at 16:26

Gabriel OlivoGabriel Olivo

63

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  A non-empty set of what?
  $endgroup$
  – José Carlos Santos
  Mar 26 at 16:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Elements from what kind of set? Totally ordered? Poset? Lattice?
  $endgroup$
  – Wuestenfux
  Mar 26 at 16:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You should add the word $finite$ to your title.
  $endgroup$
  – John Douma
  Mar 26 at 16:36
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You might need to prove that if $A$ is a finite set with $n$ elements then it can not be a finite set with $m;m ne n$ elements. Or that a set can't be both finite and infinitely countable. Or simply a set must have one specific cardinality and not two. Also twice you seem to claim elements are distinct because they are strictly greater. It's enough to say they are distinct because elements of sets are distinct by definition.
  $endgroup$
  – fleablood
  Mar 26 at 17:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can sort the numbers, so it is obvious that the set must have a minimum and a maximum (which coincide in the case the numbers are all equal)
  $endgroup$
  – Peter
  Mar 26 at 17:53
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

My proof goes as follows:

Let S be a finite non-empty set of real numbers with exactly n elements. Assume by contradiction that S does not a max value.

That is, for every element in S there exists another element in S that is strictly greater than it.

Note that these two must be distinct since the latter will be strictly greater than the former.

Now, choose any element in S, and denote it by x_1. By our assumption, there is a distinct element in S strictly greater than x_1, which we denote by x_2.

Continuing in this fashion, for x_n-1 in S, there is x_n in S strictly greater than x_n-1. However, by our assumption there will also exist x_n+1 in S strictly greater than X_n.

Since x_n+1 cannot be x_n and since S has exactly n elements, then x_n+1 must be one of the following: x_1 , x_2, ..... x_n-1. However, these are all strictly less than x_n. So, this is impossible. That is, we have a contradiction in any case.

The case for the min value of S follows similarly.

My questions are

1. Is this proof correct?




2. How do I formalize it? Assuming this isn't rigorous enough.

real-analysis

share|cite|improve this question

edited Mar 26 at 17:02

J. W. Tanner

4,9371420

asked Mar 26 at 16:26

Gabriel OlivoGabriel Olivo

63

$endgroup$

share|cite|improve this question

edited Mar 26 at 17:02

J. W. Tanner

4,9371420

asked Mar 26 at 16:26

Gabriel OlivoGabriel Olivo

63

share|cite|improve this question

edited Mar 26 at 17:02

J. W. Tanner

4,9371420

asked Mar 26 at 16:26

Gabriel OlivoGabriel Olivo

63

edited Mar 26 at 17:02

J. W. Tanner

4,9371420

edited Mar 26 at 17:02

J. W. Tanner

4,9371420

asked Mar 26 at 16:26

Gabriel OlivoGabriel Olivo

63

asked Mar 26 at 16:26

Gabriel OlivoGabriel Olivo

63