2 added 4 characters in body
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The commands \lbrack and \rbrack are macros that produce [ and ] respectively, and MathJax has no problem when you use them instead of the symbols (see- see the comment below), which was produced by

For any $a,b\in \lbrack 0,x)$: $a\vee b$ exists. Namely, $a\vee b=\max\{a,b\}$.
You might be asking about when directed sets are
[complete](http://en.wikipedia.org/wiki/Complete_partial_order)?

Incidentally, note that you should use \max instead of just max to produce a math operator.

The commands \lbrack and \rbrack are macros that produce [ and ] respectively, and MathJax has no problem when you use them instead of the symbols (see comment below), which was produced by

For any $a,b\in \lbrack 0,x)$: $a\vee b$ exists. Namely, $a\vee b=\max\{a,b\}$.
You might be asking about when directed sets are
[complete](http://en.wikipedia.org/wiki/Complete_partial_order)?

Incidentally, note that you should use \max instead of just max to produce a math operator.

The commands \lbrack and \rbrack are macros that produce [ and ] respectively, and MathJax has no problem when you use them instead of the symbols - see the comment below, which was produced by

For any $a,b\in \lbrack 0,x)$: $a\vee b$ exists. Namely, $a\vee b=\max\{a,b\}$.
You might be asking about when directed sets are
[complete](http://en.wikipedia.org/wiki/Complete_partial_order)?

Incidentally, note that you should use \max instead of just max to produce a math operator.

1
source | link

The commands \lbrack and \rbrack are macros that produce [ and ] respectively, and MathJax has no problem when you use them instead of the symbols (see comment below), which was produced by

For any $a,b\in \lbrack 0,x)$: $a\vee b$ exists. Namely, $a\vee b=\max\{a,b\}$.
You might be asking about when directed sets are
[complete](http://en.wikipedia.org/wiki/Complete_partial_order)?

Incidentally, note that you should use \max instead of just max to produce a math operator.