Linear ordering isomorphic to an initial segment Question: Suppose $\left( L, \prec \right)$ is a linear ordering such that for every $X \subseteq L, \left( X, \prec \right)$ is isomorphic to an initial segment of $\left( L, \prec \right)$. Show that $\left( L, \prec \right)$ is a well ordering. 
This is what i have gotten so far.
In order to show that $\left( L, \prec \right)$ well ordering I need to show that $\left( L, \prec \right)$ is a linear ordering and every non empty subset of $X$ has a $\prec-$ least member.
Since $\left( X, \prec \right) \cong \left( L, \prec \right)$ there is a bijective function $f:X \to L$ such that for any $x, y \in X$, if $x \prec y$ than $f(x)\prec f(y)$.
Since $f$ sends $x \in X$ to $f(x) \in L$, there is a subset $W=\{v \in L : f(v) \prec v \}$ which is non-empty. 
This is what I have gotten and I am stuck after this. Also, is my set $W$ wrong? Help is much appreciated. 

We don't have an isomorphism $f:(X,\prec) \, \to\, (L,\prec)$, only an order-preserving embedding such that the range of $f$ is an initial segment of $L$.
Now it's very easy: if we show that $L$ has a smallest element, then every initial segment of $L$ will have a smallest element, thus by the embedding being order-preserving, $X$ will have a smallest element, too.
To show that $L$ has minimum, simply take any singleton $X:=\{a\}$ with $a\in L$. By the hypothesis, $X$ is isomorphic to an initial segment of $L$, thus $f(a)$, the correspondent of $a$, must be minimal. 
