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Let $L$ be a linear order. if $A \subset L$ and $a \in L$, then $a$ is called a strict upper bound if $x least strict upper bound. Prove that if $L \not= \emptyset$, then $L$ has a least element.

I think I have a counterexample against this, namely $\mathbb{Z}$. This set has a linear order and every finite subset can be bounded with a strict upper bound. Let $X \subsetneq \mathbb{Z}$ and let $X$ be finite. Then $X$ has a maximal element $k$, then $k+1$ is the least strict upper bound for $X$. However, $\mathbb{Z}$ obviously has no least element in its total order. So how can this theorem be true?

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    In your counterexample, what if $X=\emptyset$? Does $X$ have a strict upper bound? – 2017-02-22
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    It indeed does not, but if I read the theorem correctly, it only requires that if a strict upper bound exists, then a least strict upper bound must exist. – 2017-02-22
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    Why isn't $42$ a strict upper bound for the empty set? – 2017-02-22
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    If $a$ is a strict upper bound for $A,$ and if $B$ is a subset of $A,$, then is $a$ also a strict upper bound for $B$? – 2017-02-22
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    Every element of $\mathbb{Z}$ is a strict upper bound for the empty set. But the empty set does not have a least strict upper bound (in this case this is a least element of $\mathbb{Z}$), hence $\mathbb{Z}$ does not satisfy the condition of the theorem. – 2017-02-22
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    Ah yes, thank you! I see the problem there – 2017-02-22

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Let $A$ be the empty subset of $L$. Any element of $L$ is a strict upper bound for $A$ since our condition becomes empty. Since $L$ is empty, $A$ has a strict upper bound, hence a least strict upper bound, which is a least element.

In your example it goes wrong for the empty set (see comments above).