Embedding of ordered sets?

Question

Let $(E,<)$ a well-ordered set. Let $g\colon(E,{<})\to(E,<)$ an embedding of ordered sets.

I have to show that $\forall e\in E, e\le g(e)$.

The hint is to consider the set $C=\{\,e\in E \mid e>g(e)\,\}$ but I don't understand what informations it gives.

Then how many embeddings there are in $(E,<)$ ? Just the trivial one ?

Thanks in advance !

Answer 1

HINT: If the theorem were false, the set $C$ would be a non-empty subset of $E$ and would therefore have a least element $c$ with respect to the well-order $<$. Use $c$ to get a contradiction by showing that $g$ is not order-preserving.

$E$ can have many embeddings. Take $E=\Bbb N$. Then for each $k\in\Bbb N$ the map $n\mapsto n+k$ is an embedding, for instance. In fact there are $2^\omega=\mathfrak{c}=|\wp(\Bbb N)|$ embeddings of $\Bbb N$ into itself, since for each infinite $A\subseteq\Bbb N$ there is an embedding $f_A:\Bbb N\to\Bbb N$ such that $f_A[\Bbb N]=A$.

Answer 2

The hint is to consider $C$. Note that you want to show that $C=\emptyset$. So assume $C\ne\emptyset$. What can be said about non-empty subsets of a well-ordered sets? What can you conclude?

Re your second question: $x\mapsto 7x^2+5$ is a non-trivial embedding $(\Bbb N,<)\to (\Bbb N,<)$.