# Metric Space

• Jan 10th 2009, 04:53 PM
aliceinwonderland
Metric Space
Show that the limit of a convergent sequence of distinct points in a metric space is a limit point of the range of the sequence. Give an example to show that this is not true if the word "distinct" is omitted.
• Jan 10th 2009, 06:37 PM
ThePerfectHacker
Quote:

Originally Posted by aliceinwonderland
Show that the limit of a convergent sequence of distinct points in a metric space is a limit point of the range of the sequence. Give an example to show that this is not true if the word "distinct" is omitted.

Let $\{x_n\}$ is a sequence. And $X$ be the range of the sequence. Now a point $x\in X$ is a limit point if for any $r > 0$ we have that $B(x,r)\cap X$ has infinitely many points. Thus, we see that if $x = \lim x_n$ is a limit point because for any $r > 0$ pick $\epsilon = r$ and so $d(x_n,y) < \epsilon$ for $n\geq N$ where $N$ is a natural number. Thus, $\{x_N,x_{N+1},x_{N+2}, ... \}$ is contained in $B(x,r)\cap X$. However, this set is infinite since $\{x_n\}$ is a sequence of distinct points.

For a counter example think of konstant sequences.
• Jan 10th 2009, 09:50 PM
archidi
Mr perfect Hacker and aliceinwonderland YOU are both mistaken in the definition of x being an accumulation point of the sequence { $\ x_{n}$}

SERGE LANG in his book analysis I ,and on page 32 ,comes to justify me on that >

He writes and i quote:

"LET { $\ x_{n}$} be a sequence and x a No.We shall say that x is apoint of accumulation (or limit pt) of the sequence if given ε there exists infinitely many integers n such that:
.............................................| $\ x_{n}-x$|<ε............................................... ..........................................

AND then he goes on to give couple of examples:

1) The sequence {1,1,1,.............} has one point of accumulation namely 1

2) The sequence {1,1/2,1,1/3,1,1/4............}has two points of accumulations,namely 1 and 0.

3) The sequence { 1,2,3,.................} has no point of accumulation."

So the word "distinct" is of no importance as the above definition and examples clearly show

Another working definition , since the phrase infinitely many is of no substance in a proof, is the following:

x is a limit point of $\ x_{n}$ iff for all ε>0 and forall kεN ( =k belonging to the natural Nos) there exists a natural No $m\geq k$ such that:

...................................| $\ x_{m}-x$|<ε............................................... ...............................................

Now based on that definition i will try to give a proof that:

If $lim\ x_{n} = x$ ,then, for all ε>0 and forall kεN ( =k belonging to the natural Nos) there exists a natural No $m\geq k$ such that:

...................................| $\ x_{m}-x$|<ε............................................... ...............................................

Let ε>0 and kεΝ,now since $lim\ x_{n} = x$ there exists a natural No r ,rεΝ such that.

..............................| $\ x_{n}-x$|<ε...for all n $\geq r$.................................................. .....................

And now we distinguish two cases :

CASE 1 $k\geq r$.

CASE 2 r>k.

IN case 1 we choose $m\geq k$ and in case we choose $m\geq r$

Considering now case 1 ,let $m\geq k$ then we have $m\geq r$ and therefor .

................................| $\ x_{m}-x$|<ε............................................... ..............................................

NOTE the above proof was done in the metric space of real Nos .It can be generalized in any metric space ( X,d) by substituting the absolute value in real Nos by the general metric ,d,and the word, No by the word ,point

For e,g | $\ x_{n}-x$| by $d(\ x_{n},x)$
• Jan 10th 2009, 11:58 PM
aliceinwonderland
Quote:

Originally Posted by archidi
1) The sequence {1,1,1,.............} has one point of accumulation namely 1

I think if the sequence A = {1,1,1,.............} (I assume no element except 1 is in A) is in a metric space, 1 is not a limit point of A. Since any open set containing 1 should intesect A distinct from 1, which is impossible.
• Jan 11th 2009, 04:35 AM
kalagota
well, the two definitions you gave are indeed equivalent. but did you read the question carefully? what do you think they wanted to prove?

next,

Quote:

Originally Posted by archidi
Now based on that definition i will try to give a proof that:

If $lim\ x_{n} = x$ ,then, for all ε>0 and forall kεN ( =k belonging to the natural Nos) there exists a natural No $m\geq k$ such that:

...................................| $\ x_{m}-x$|<ε............................................... ...............................................

why do you have to prove this? this is just a part of the definition. i mean, there is nothing to prove in here.
• Jan 11th 2009, 04:47 AM
HallsofIvy
Quote:

Originally Posted by archidi
Mr perfect Hacker and aliceinwonderland YOU are both mistaken in the definition of x being an accumulation point of the sequence { $\ x_{n}$}

But the problem was about limit points, not accumulation points. Neither thePerfectHacker nor aliceinwonderland said anything about "accumulation points".