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 $\displaystyle \{x_n\}$ is a sequence. And $\displaystyle X$ be the range of the sequence. Now a point $\displaystyle x\in X$ is a limit point if for any $\displaystyle r > 0$ we have that $\displaystyle B(x,r)\cap X$ has infinitely many points. Thus, we see that if $\displaystyle x = \lim x_n$ is a limit point because for any $\displaystyle r > 0$ pick $\displaystyle \epsilon = r$ and so $\displaystyle d(x_n,y) < \epsilon$ for $\displaystyle n\geq N$ where $\displaystyle N$ is a natural number. Thus, $\displaystyle \{x_N,x_{N+1},x_{N+2}, ... \}$ is contained in $\displaystyle B(x,r)\cap X$. However, this set is infinite since $\displaystyle \{x_n\}$ 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 {$\displaystyle \ x_{n}$}
SERGE LANG in his book analysis I ,and on page 32 ,comes to justify me on that >
He writes and i quote:
"LET {$\displaystyle \ x_{n}$} 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:
.............................................|$\displaystyle \ x_{n}-x$|<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 $\displaystyle \ x_{n}$ iff for all ε>0 and forall kεN ( =k belonging to the natural Nos) there exists a natural No $\displaystyle m\geq k$ such that:
...................................|$\displaystyle \ x_{m}-x$|<ε............................................... ...............................................
Now based on that definition i will try to give a proof that:
If $\displaystyle lim\ x_{n} = x$ ,then, for all ε>0 and forall kεN ( =k belonging to the natural Nos) there exists a natural No $\displaystyle m\geq k$ such that:
...................................|$\displaystyle \ x_{m}-x$|<ε............................................... ...............................................
Let ε>0 and kεΝ,now since $\displaystyle lim\ x_{n} = x$ there exists a natural No r ,rεΝ such that.
..............................|$\displaystyle \ x_{n}-x$|<ε...for all n$\displaystyle \geq r$.................................................. .....................
And now we distinguish two cases :
CASE 1 $\displaystyle k\geq r$.
CASE 2 r>k.
IN case 1 we choose $\displaystyle m\geq k$ and in case we choose $\displaystyle m\geq r$
Considering now case 1 ,let $\displaystyle m\geq k$ then we have $\displaystyle m\geq r$ and therefor .
................................|$\displaystyle \ x_{m}-x$|<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 |$\displaystyle \ x_{n}-x$| by $\displaystyle d(\ x_{n},x)$