Equal cardinality implies isomorphism of ordering I think I have proved this (should-be false) lemma.

For any set $X$, if $X$ is equipotent to an ordinal $\alpha$, then $X$ can be ordered so that it is isomorphic to $\alpha$.

This is the proof

Let $X$ be a set such that it is equipotent to an ordinal $\alpha$. Thus there exists a bijection $f:X\rightarrow\alpha$. We then define the set $\prec$ to be
      \begin{equation*}
 \prec:=\lbrace(x,y)\in X\times X:f(x)<f(y)\rbrace.
 \end{equation*}
  To prove that $\prec$ is really a strict ordering, we are required to show that $\prec$ is asymmetric and transitive. Firstly, to prove that it is asymmetric, we suppose that $x\prec y$ holds. We assume for the sake of contradiction that $y\prec x$ also holds. Thus $f(x)<f(y)$ and $f(y)<f(x)$ which is a contradiction. Thus $y\prec x$ does not hold and hence $\prec$ is asymmetric. Secondly, to prove that $\prec$ is transitive, we suppose that $x\prec y$ and $y\prec z$. Thus $f(x)<f(y)$ and $f(y)<f(z)$. Since $<$ is transitive, it follows that $f(x)<f(z)$. Since $x,z\in X$, this implies that $x\prec z$. Thus $\prec$ is transitive and indeed a strict ordering.
      Now, if $x,y\in X$ and $x\prec y$, then $f(x)<f(y)$ and thus $f$ is the isomorphism between $X$ and $\alpha$.

The problem is, if this lemma is true, then this creates a lot of contradiction. For example, $\omega=\omega+1$. This is because $|\omega|=|\omega+1|$ and so (from this lemma) $\omega$ is isomorphic to $\omega+1$. But since no well-ordered set can be isomorphic to its own initial segment, it follows that $\omega=\omega+1$.
Something here is wrong! 
Thank you in advance

Can be ordered is right. The ordering need not agree with another ordering on $X$.  There is no contradiction.
