# Clarification needed

## Homework Statement

Prove that a non-empty set of complex numbers F is closed iff every convergent sequence of elements of F converges to an element of F.

## The Attempt at a Solution

I don't understand the second part of the iff statement. If someone could clarify what it means about convergent sequences, I could go from there.

## Answers and Replies

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LCKurtz
Homework Helper
Gold Member

## Homework Statement

Prove that a non-empty set of complex numbers F is closed iff every convergent sequence of elements of F converges to an element of F.

## The Attempt at a Solution

I don't understand the second part of the iff statement. If someone could clarify what it means about convergent sequences, I could go from there.
Not sure what you call the "second part". You have to show two statements:

1. If every convergent sequence of elements of F converges to an element of F, then F is closed.

2. If F is closed, then every convergent sequence of elements in F converges to an element of F.

I don't understand what "every convergent sequence of elements of F converges to an element of F" means.

Bacle2
Well, {1/n} n=1,.... is a sequence of elements in the interval (0,1), and the

sequence converges to ___ which is in ____.......

it converges to 0, which is in the interval, or the border of the interval. But 0 isn't an element of that sequence though.

Bacle2
Actually, 0 is _not_ in the interval. Remember:

(0,1):={ x: 0<x<1 } , which does not include 0 . Tho this is just for the

real line.

So how does that example work in this situation then. We needed a sequence who's limit is an element of the sequence.

Bacle2
I was trying to illustrate a familiar case in which a sequence defined on S does not

converge to a point in S. What is your book's definition of closed set ?

We don't have a book, but basically a closed set is one where the boundary of the set is contained in the set.

Bacle2
Maybe this is the nicest , more general way of doing things:

If F is closed in ℂ , then U:= ℂ\F is open . Since U is ( U are? ) open, if x is in U ,

then there is a 'hood ( neighborhood) of x contained in U , i.e., ( in this context) a

ball B(x,r) contained in U . What would happen if x was a limit point of some sequence

{fn} in F , could you find such a ball ?

Bacle2
Sorry, LCKurtz, I did not mean to strong-arm myself into the post and take-it over.

LCKurtz
Homework Helper
Gold Member
Sorry, LCKurtz, I did not mean to strong-arm myself into the post and take-it over.
No problem. I have been gone all afternoon anyway. And judging from the responses, it may take both of us to explain the concept.

LCKurtz
F is a subset. Every convergent sequence in F has a limit. This limit may or may not be in F. Look at Bacle's example on the real line. If F = (0,1) and $x_n=\frac 1 {n+1}$ then $x_n\rightarrow 0$, which is not in F. So F isn't closed. Such limits must all be in F if F is to be closed. If F = [0,1], then it is closed, as an example.