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I managed to quite easily show the reverse direction $\Leftarrow$, but the following direction is giving me a lot of problems:

no strictly decreasing sequence of elements in linear order $A$ $\Rightarrow$ $A$ is well ordered

Any help would be appreciated!

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marked as duplicate by ℋolo, Carl Mummert logic Apr 27 at 16:55

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Suppose $A$ is not well-ordered by $<$. Then there is a non-empty subset $B$ of $A$ without a minimum.

So pick $x_0 \in B$. Then there is $x_1 \in B$ with $x_1 < x_0$ (or else $x_0 = \min(B)$).

Having picked $x_n \in B$ such that $x_n < x_{n-1} < \ldots < x_0$ we can pick $x_{n+1} \in B$ such that $x_{n+1} < x_n$ (or $x_n$ would have been the minimum of $B$, but this does not exist).

So by recursion we have defined a descreasing strictly sequence $(x_n)$ in $A$.

This shows the other direction by contrapositive (not a well-order implies the existence of a strictly decreasing sequence).

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  • $\begingroup$ Thank you very much! Accepted. $\endgroup$ – nshct Apr 27 at 22:30

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