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 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.
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