What is the relationship between the limit and supremum of a sequence?

[STEMucator]

Homework Statement

http://gyazo.com/d59c730eb9b18dda4504a5fe118c7213

Homework Equations

Limit and supremum.

The Attempt at a Solution

(a) Let : $b_n = a_n - b$ so that $b_n ≤ 0$

Now, $lim(b_n) = lim(a_n - b) ≤ 0 \Rightarrow a - b ≤ 0 \Rightarrow a ≤ b$

Q.E.D

(b) I'm actually having some trouble with this one. I want to show the limit of the sequence $a_n$ is less than or equal to the supremum of the sequence $a_n$ while $n$ varies over $\mathbb{N}$. Where do I start with this one?

 

[christoff]

Part a) is valid, but only if you know (or could prove) that the limit of a non-positive convergent sequence is non-positive.

As for part b), there are a couple of ways you could do it. My idea is following: First, there are two simple cases:
1) Your sequence is decreasing.
2) Your sequence is increasing.

In case 1, we have $sup_na_n=a_1$ since the sequence is decreasing, so you trivially have $a\leq sup_na_n=a_1$. The situation is similar for part 2), and in fact can be seen via case 1 by considering the sequence $b_n=-a_n$ and properties of limits of sequences and multiplication by constants (check this!).

In the general case, can you somehow put these two situations together by "splitting" your sequence into two subsequences, one in which all terms are smaller than or equal to the limit, and another where all of the terms are larger than the limit? What do you know about subsequences of convergent sequences?

 

[STEMucator]

Part a) is valid, but only if you know (or could prove) that the limit of a non-positive convergent sequence is non-positive.

As for part b), there are a couple of ways you could do it. My idea is following: First, there are two simple cases:
1) Your sequence is decreasing.
2) Your sequence is increasing.

In case 1, we have $sup_na_n=a_1$ since the sequence is decreasing, so you trivially have $a\leq sup_na_n=a_1$. The situation is similar for part 2), and in fact can be seen via case 1 by considering the sequence $b_n=-a_n$ and properties of limits of sequences and multiplication by constants (check this!).

In the general case, can you somehow put these two situations together by "splitting" your sequence into two subsequences, one in which all terms are smaller than or equal to the limit, and another where all of the terms are larger than the limit? What do you know about subsequences of convergent sequences?

Ahhh I see, so in case 2 I would say consider the sequence $b_n = -a_n$ which is the sequence from part (1), but it is now increasing. So I can conclude that $a ≤ sup_n a_n = -a_1$ in this case.

Yes if I broke the sequence into sub sequences, let's say $\forall n \in \mathbb{N}$ :

$u_n = \{ x \in a_n \space | \space x ≤ a \}$
$v_n = \{ x \in a_n \space | \space x > a \}$

I know every sub sequence of a convergent sequence converges, which would imply $u_n$ and $v_n$ are convergent so that $u_n + v_n$ is also convergent.

 

[christoff]

Ahhh I see, so in case 2 I would say consider the sequence $b_n = -a_n$ which is the sequence from part (1), but it is now increasing. So I can conclude that $a ≤ sup_n a_n = -a_1$ in this case.

Not quite. Also , it might not be as easy as I originally thought. Showing it explicitly isn't too hard though... Once again, contradiction is the way to go.

Suppose $a_n$ is increasing and (by assuming the false result),$sup_na_n:=s_A<a$. But then since $a$ is the limit of the sequence, there exists $N>0$ such that $|a-a_N|<a-s_A$ (ie. in the epsilon definition of limit, we take $\epsilon=a-s_A$). But then we have $a_N>s_A$, a contradiction. The case 2) for increasing sequences is therefore proven; $sup_na_n\geq a$.

Yes if I broke the sequence into sub sequences, let's say ∀n∈N :

un={x∈an | x≤a}
vn={x∈an | x>a}

I know every sub sequence of a convergent sequence converges, which would imply un and vn are convergent so that un+vn is also convergent.

That's almost it, but be careful. It isn't a very good idea to think of those two new sequences as being something that should be added together. The reason for this is that although $u_n+v_n$ is convergent the way you've defined it, the limit is actually equal to $2a$, where $a$ is the original limit.

A better way to think about the problem is this: define the set $U=\{u_n:n\in\mathbb{N}\}$. So U is the set of all points in your sequence $u_n$. Define $V$ similarly for the sequence $v_n$. So then $sup_nu_n=sup(U)\leq a$, and $sup(V)\leq a$ by the properties above for increasing and decreasing sequences. Now, note that $U\cup V=\{a_n:n\in\mathbb{N}\}$, so that $sup_na_n=sup(U\cup V)$. Do you know any useful properties about the supremum of a union of two sets?

The next step would be to show that: for any two (bounded) sets $A,B$, we have $sup(A\cup B)\leq max\{sup(A),sup(B)\}$, then you would be done, since then no matter which had the larger supremum (of U and V, in the problem we're solving), both are still less than $a$.

(In fact, you can show $sup(A\cup B) = max\{sup(A),sup(B)\}$, but for this proof, it isn't necessary).

 

[STEMucator]

Not quite. Also , it might not be as easy as I originally thought. Showing it explicitly isn't too hard though... Once again, contradiction is the way to go.

Suppose $a_n$ is increasing and (by assuming the false result),$sup_na_n:=s_A<a$. But then since $a$ is the limit of the sequence, there exists $N>0$ such that $|a-a_N|<a-s_A$ (ie. in the epsilon definition of limit, we take $\epsilon=a-s_A$). But then we have $a_N>s_A$, a contradiction. The case 2) for increasing sequences is therefore proven; $sup_na_n\geq a$.



That's almost it, but be careful. It isn't a very good idea to think of those two new sequences as being something that should be added together. The reason for this is that although $u_n+v_n$ is convergent the way you've defined it, the limit is actually equal to $2a$, where $a$ is the original limit.

A better way to think about the problem is this: define the set $U=\{u_n:n\in\mathbb{N}\}$. So U is the set of all points in your sequence $u_n$. Define $V$ similarly for the sequence $v_n$. So then $sup_nu_n=sup(U)\leq a$, and $sup(V)\leq a$ by the properties above for increasing and decreasing sequences. Now, note that $U\cup V=\{a_n:n\in\mathbb{N}\}$, so that $sup_na_n=sup(U\cup V)$. Do you know any useful properties about the supremum of a union of two sets?

The next step would be to show that: for any two (bounded) sets $A,B$, we have $sup(A\cup B)\leq max\{sup(A),sup(B)\}$, then you would be done, since then no matter which had the larger supremum (of U and V, in the problem we're solving), both are still less than $a$.

(In fact, you can show $sup(A\cup B) = max\{sup(A),sup(B)\}$, but for this proof, it isn't necessary).

Yes indeed, $sup_n a_n = sup(U \cup V) ≤ max\{sup(U), sup(V)\} = a$ in this case.

 

[christoff]

Correct. On a purely technical note however, that last equals sign should be a $\leq$, since you technically don't know what either of sup(U) or sup(V) is, only that they are both bounded by a (so less than or equal to a).

 

[Bacle2]

Another approach: If you had a>b , then consider, e.g., (a-b)/2 . Since a is the limit of {a_n} , there should be

some point(s) ( actually, infinitely-many) between a and b , but this is not possible...

Something similar for second part.