# Prove if there is no strictly decreasing sequence of elements in linear order A, then A is well ordered [duplicate]

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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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).

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