1. ## 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.

2. 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.

3. 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 $d(\ x_{n},x)$

4. 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.

5. 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,

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.

6. 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".