I am trying to say because there are infinite members of the sequence in the limit L area, then we should look at the first memeber in L-1, then we should look at the member in area L-1/2, first member is bigger than the first one, lets assume its not then then there wont be infinite members in area nearer tban that near L so its a contradiction,
All this is great but how mathmatically I write it?
You don't need the sequence to be converging at all. In any linearly ordered set every sequence has an increasing or a decreasing subsequence. For the classical argument for this, see this answer.
If you insist on making use of the convergence: there are either infinitely many members of the sequence $<L$ or infinitely many $>L$. Pass to the subsequence of only those elements, say we have all that are $<L$. Then pick $x_{n_1}$ in that subsequence. Then as the subsequence converges to $L$, there is a sequence element $x_{n_2} > x_{n_1}$ (use $\varepsilon = L - x_{n_1} > 0$ in the definition of convergence). Proceed by recursion.
This last argument can also be used for the general real-number case: use a 2-point ordered compactification and pass to a convergent subsequence first....
Here's a proof using the limit inferior of a sequence. I suppose this argument can (after some modification) be extended to any arbitrary sequence (whether convergent or not).
Suppose $\{x_{n}\}_{n=1}^{\infty}$ is convergent. Let $\lim_{n\to\infty} x_{n} = \ell$. Consider $y_{n} = \inf_{k \geq n} \{x_{k}\}$.
Case 1: If there exists $n_{0} \in \mathbb{N}$ such that $y_{n_{0}} = \ell$ then by definition of infimum on the set $\{ x_{n_{0}}, x_{n_{0}+1}, x_{n_{0}+2}, \dots \}$ we have $x_{k} \geq \ell$ for all $k \geq n_{0}$. We claim that this allows us to choose a monotonically decreasing subsequence of $\{x_{n}\}_{n=n_{0}}^{\infty}$. Suppose not, then there exists $n_{1} \geq n_{0}$ such that whenever $m \geq n_{1}+1 > n_{1}$ we have $x_{m} \geq x_{n_{1}+1} > x_{n_{1}} \geq \ell$. This implies that $\lim_{m\to\infty} x_{m}= \ell \geq x_{n_{1}+1} > x_{n_{1}} \geq \ell$, which is a contradiction.
Case 2: For all $n \in \mathbb{N}$ we have $y_{n} \neq \ell$. It is clear that we must then have $y_{n} < \ell$ for all $n \in \mathbb{N}$, for if we had $y_{n} > \ell$ for some $n \in \mathbb{N}$, this would imply that for all $k \geq n$ we have $x_{k} \geq y_{n} > \ell$ and hence $\lim_{n \to \infty} x_{n} = \ell \geq y_{n} > \ell$, another contradiction. Now, a useful property of $\{ y_{n} \}_{n=1}^{\infty}$ is that it is monotonically increasing (why?). We then claim that for each $n \in \mathbb{N}$, there exists $m \geq n$ such that $y_{n} = \inf_{k\geq n}\{x_{k}\} = x_{m}$. If this was not true, then for all $k > n$ we have $x_{k} > \inf_{k\geq n} \{x_{k}\}$ which implies that for any $\varepsilon > 0$ we can always find a $k \in \mathbb{N}$ such that $0 < x_{k} - y_{n} < \varepsilon$. By observing $x_{n} - y_{n}$ followed by the minimal $k_{1} > n$ such that $x_{k_{1}} - y_{n} < \frac{1}{2}(x_{n} - y_{n})$, and (inductively) the minimal $k_{i+1} > k_{i}$ such that $x_{k_{i+1}} - y_{n} < \frac{1}{2^{i}}(x_{k_{i}} - y_{n})$ leads to the conclusion that there is a subsequence $\{x_{k_{i}}\}_{i=1}^{\infty}$ converging to $y_{n} \neq \ell$, contradicting our assumption that $\lim_{n\to\infty} x_{n} = \ell$. Therefore we may define a monotonically increasing subsequence as follows. Choose $x_{m_{1}} = \inf_{k\geq 1} \{ x_{k} \} = y_{1}$. Subsequently (pun intended, ha ha) for each $i \in \mathbb{N}$ choose $x_{m_{i+1}} = \inf_{k \geq m_{i} + 1} \{x_{k}\} = y_{m_{i}+1}$ and we are done.
