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.

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- Jan 10th 2009, 04:53 PMaliceinwonderlandMetric 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 PMThePerfectHacker
Let is a sequence. And be the range of the sequence. Now a point is a limit point if for any we have that has infinitely many points. Thus, we see that if is a limit point because for any pick and so for where is a natural number. Thus, is contained in . However, this set is infinite since is a sequence of distinct points.

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

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

He writes and i quote:

"LET { } 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:

.............................................| |<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**iff**for all ε>0 and forall kεN ( =k belonging to the natural Nos) there exists a natural No such that:

...................................| |<ε............................................... ...............................................

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

If ,**then,**for all ε>0 and forall kεN ( =k belonging to the natural Nos) there exists a natural No such that:

...................................| |<ε............................................... ...............................................

Let ε>0 and kεΝ,now since there exists a natural No r ,rεΝ such that.

..............................| |<ε...for all n .................................................. .....................

And now we distinguish two cases :

CASE 1 .

CASE 2 r>k.

IN case 1 we choose and in case we choose

Considering now case 1 ,let then we have and therefor .

................................| |<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 | | by - Jan 10th 2009, 11:58 PMaliceinwonderland
- Jan 11th 2009, 04:35 AMkalagota
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