# Prove that a sequence of subsequential limits contains inf and sup (1 Viewer)

### Users Who Are Viewing This Thread (Users: 0, Guests: 1)

#### Sick0Fant

Okay. The problem I have is:

Let {x_n} be bdd and let E be the set of subsequential limits of {x_n}. Prove that E is bdd and E contains both its lowest upper bound and its greatest lower bound.

So far, I have:
{x_n} is bdd => no subseq of {x_n} can converge outside of {x_n}'s bounds=>E is bounded.
Now, sse that y=sup(E) is not in E=> there is a z in E s.t. y-e < z < y for some e > 0.

Now, how would one proceed from here?

#### mathman

You can generate a sequence of z's by using a sequence of e's that goes to 0. This sequence of z's the must converge to y.

#### Sick0Fant

I already had thought of that: you have y - e< z < y. Take e to be 1/k with e going to infinity, then {z_k} cgt to y, but what can we really conclude from that? Is there any guarentee that a {z_k} is in the original seq?

#### mathman

If you can't find a z for any e>0, then y is > sup(E)

### The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving