# Thread: Help with Real Analysis

1. ## Help with Real Analysis

Hello,

I have two homework problems which I am really struggling to grasp conceptually. I was unsure how to attach images, so I have uploaded a picture of the problems from the book to imgur, https://imgur.com/a/cK7Eo. The first is question 3, and the second is question 5 (circled). Please note that the signs in problem 3 which appear to be strict subsets are just regular subsets, there does not need to be a proof that they are strict. Here are my proofs so far, any assistance would really help as I am very unsure of what I am dealing with here. Thank you so much (and sorry about the image quality)! (The book is Berberian's A First Course in Real Analysis) (I can't get the formatting to work properly, it's putting weird indents all over the place. Sorry about that.)

The definitions we were supposed to use:
Ultimately: We say that xn in A ultimately if xn belongs to A for some index onward. That is, there is an index N such that xn in A for all n greater than or equal to N.
Frequently: We say that xn in A frequently if for every index N there is an index n greater than or equal to N for which xn in A.

3. Proof. Let An be a sequence of subsets of a set X. Define
limsup(An)={x in X : x in An frequently}
liminf(An)={x in X : x in An ultimately}
(i) By Thm 3.2.4, only one of the following conditions hold:
(1) xn in A ultimately
(2) xn not in A frequently.
Then since (1) is true, this implies xn in A frequently. Thus,
liminf(An) is a subset of limsup(An).
(ii - I didn't really understand the notation or what it meant, I basically just stated the definition due to this) By the definition of ultimately since
liminf(An)={x in X : x in An ultimately}, this implies that there exists an n in Z>0 such that for an index k greater than or equal to n, there is an xk in Ak. Then liminf(An)=U n Ak. (This last part is the statement in the question, see the image, I am not sure how to use LaTeX).
(iii) (
limsup(An))'={x in X : x not in An frequently}. Then by Thm 3.2.4 previously stated, (2) holds. This implies xn not in An ultimately, which is equivalent to {x in X : x not in An ultimately}=liminf(An)'.
(iv) Since
limsup(An)={x in X : x in An frequently}, by the definition of frequently this implies that for all n in Z>0 there exists an index k greater than or equal to n such that xk in Ak. Then
limsup(An)=n U Ak. (Again, this is the statement in the image of iv)

5. First, let me give the definition of null as the book provides it: A sequence
(xn) in R is said to be null if, for every positive real number ɛ, |(xn)|<ɛ, ultimately.
Proof. Let (xn) in R be a sequence.
(=>) Suppose
(xn) is not null. Then by the negating the definition of a null sequence, we have there exists ɛ such that |xn| is greater than or equal to ɛ frequently.
(<=) Suppose there exists
ɛ such that |xn| is greater than or equal to ɛ frequently. Then suppose (xn) is null. Then by the definition of null,(xn) is greater than ɛ ultimately for all ɛ in Z>0.
This is a contradiction, so the assumption that
(xn) is null must be false. Then (xn) is not null.

2. ## Re: Help with Real Analysis

Originally Posted by Barbs
I have two homework problems which I am really struggling to grasp conceptually. I was unsure how to attach images, so I have uploaded a picture of the problems from the book to imgur, https://imgur.com/a/cK7Eo. The first is question 3, and the second is question 5 (circled). Please note that the signs in problem 3 which appear to be strict subsets are just regular subsets, there does not need to be a proof that they are strict. Here are my proofs so far, any assistance would really help as I am very unsure of what I am dealing with here. Thank you so much (and sorry about the image quality)! (The book is Berberian's A First Course in Real Analysis) (I can't get the formatting to work properly, it's putting weird indents all over the place. Sorry about that.)
The definitions we were supposed to use:
Ultimately: We say that xn in A ultimately if xn belongs to A for some index onward. That is, there is an index N such that xn in A for all n greater than or equal to N.
Frequently: We say that xn in A frequently if for every index N there is an index n greater than or equal to N for which xn in A.

3. Proof. Let An be a sequence of subsets of a set X. Define
limsup(An)={x in X : x in An frequently}
liminf(An)={x in X : x in An ultimately}
(i) By Thm 3.2.4, only one of the following conditions hold:
(1) xn in A ultimately
(2) xn not in A frequently.
Then since (1) is true, this implies xn in A frequently. Thus,
liminf(An) is a subset of limsup(An).
(ii - I didn't really understand the notation or what it meant, I basically just stated the definition due to this) By the definition of ultimately since
liminf(An)={x in X : x in An ultimately}, this implies that there exists an n in Z>0 such that for an index k greater than or equal to n, there is an xk in Ak. Then liminf(An)=U n Ak. (This last part is the statement in the question, see the image, I am not sure how to use LaTeX).
(iii) (
limsup(An))'={x in X : x not in An frequently}. Then by Thm 3.2.4 previously stated, (2) holds. This implies xn not in An ultimately, which is equivalent to {x in X : x not in An ultimately}=liminf(An)'.
(iv) Since
limsup(An)={x in X : x in An frequently}, by the definition of frequently this implies that for all n in Z>0 there exists an index k greater than or equal to n such that xk in Ak. Then
limsup(An)=n U Ak. (Again, this is the statement in the image of iv)

5. First, let me give the definition of null as the book provides it: A sequence
(xn) in R is said to be null if, for every positive real number ɛ, |(xn)|<ɛ, ultimately.
Proof. Let (xn) in R be a sequence.
(=>) Suppose
(xn) is not null. Then by the negating the definition of a null sequence, we have there exists ɛ such that |xn| is greater than or equal to ɛ frequently.
(<=) Suppose there exists
ɛ such that |xn| is greater than or equal to ɛ frequently. Then suppose (xn) is null. Then by the definition of null,(xn) is greater than ɛ ultimately for all ɛ in Z>0.
This is a contradiction, so the assumption that
(xn) is null must be false. Then (xn) is not null.
My Norton goes wild if I try to look at your link. I also will say that those of us who are in the Moore/Lane tradition think of Berberian as a barbarian. (that is an inside joke at UT). If you mean that you cannot use LaTeX coding here, it must be raped in \$\$ delimiters.

I suggest that you repost the question in exact language.

3. ## Re: Help with Real Analysis

Question 3: Let An be a sequence of subsets of a set X. Define
limsup(An)={x in X : x in An frequently}
liminf(An)={x in X : x in An ultimately}.
Prove:
(i) liminf(An) is a subset of limsup(An).
(ii) liminf(An)= U (n=1 to infinity) ∩ (n=k to infinity) Ak.
(iii) (limsupAn)'=liminfAn' (' means complement).
(iv) limsup(An)= ∩ (n=1 to infinity) U (n=k to infinity) Ak.

Question 5: Show that a sequence (xn) in R is not null if and only if there exists a positive number ɛ such that |(xn)>/ɛ frequently. (Greater than or equal to.)

4. ## Re: Help with Real Analysis

It was just big.