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Show that a set is finite if and only if every linear ordering on it is a well-ordering

What i have done so far:

$\Rightarrow$: Let $(X,\prec)$, where X is finite and $\prec$ is linear ordering. Consider a subset $A\subset X$. Since $X$ is finite, $A$ is also finite. Hence, there exists some $f:I(x)\rightarrow A$ bijective, where $I(x)$ is a finite subset of the natural numbers. We redefine $f$ in a such way that it is an increasing function. Given $a,b\in A$, we have that $f(m)=a$ and $f(n)=b$, for some $m,n\in I(x)$. Since $\prec$ is linear ordering we have that $a\prec b$ or $b\prec a$. Considering the firt case $a\prec b$ (ie, $f(m)\prec f(n)$), we redefine $f$ only if $n

That's what i could do. Is it correct? Any hints on the $\Leftarrow$ part?

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    For $\Leftarrow$: If $\prec$ is a linear order on $X$, then $\succ$ is also a linear order. By hypothesis they are both well-orderings. You can try to show that only finite sets satisfy this. – 2017-02-21
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    Also, you can simplify your proof by using the following characterisation of well-orderings: there exists no infinite strictly decreasing sequence. – 2017-02-21
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    The characterisation of well-orderings above depends on the [axiom of dependent choice](https://en.wikipedia.org/wiki/Axiom_of_dependent_choice). – 2017-02-21

2 Answers 2

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Some axiom of choice is needed, since it is possible that there are sets which cannot be linearly ordered (and therefore must also be infinite).

But once you have the axiom of choice, every infinite set has a countably infinite subset. Now think, what sort of linear ordering can be made with a set which has a countably infinite subset and it is not a well-ordering?

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    What i tought so far: in order to prove that a linear ordering $\prec$ is not a well-ordering, we must exibit a subset wich has no minimal element. So, considering $X$ infinite and $V\subset X$ countably infinite, we define an ordering $<$ on it by putting $f(x_1)$f(x_1),f(x_2)\in V$ and $x_1,x_2\in N$. Supposing that $V$ has a minimal element $b=f(m)$, then $b$f(x)\in V$ would imply $x$x\in N$. Hence this would imply that $N$ is limited, which is an absurd. – 2017-02-22
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    You still need to linearly order the rest of $X$, though. – 2017-02-22
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For the opposite direction assume that $X$ is infinite and that any linear order on it is a well-order. Let's try to build an infinite descendant chain. Choose a linear order $<$ on $X$, the by hypothesis it is a well-order. Take a finite set $W_0$ let $a_0\in W_0$ be the bottom element according to $<$. Since $X$ is infinite you can choose another finite subset $W_1$ of $X$ disjoint from $W_0$. Let $a_1$ be the bottom of $W_1$ according to $<$, Now compare $a_0$ and $a_1$. If $a_0

Continuing this way you build a sequence $a_n$ which is an infinite descending chain.