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I tried to leave the following comment, and had a problem:

For any $a,b\in [0,x)$: $a\vee b$ exists. Namely, $a\vee b=max\{a,b\}$. You might be asking about when directed sets are complete?

It works in this question, but it seemed like in the comment code it was using the [ from the interval as the start of the link, instead of just around [complete].

I have left the same text as a comment to this question as a demonstration of the problem.

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  • $\begingroup$ For any $a,b\in [0,x)$: $a\vee b$ exists. Namely, $a\vee b=max\{a,b\}$. You might be asking about when directed sets are complete? $\endgroup$ – Xodarap Oct 9 '13 at 18:04
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    $\begingroup$ Seems like the interval [0,x) is causing trouble. Should the interval be [0,x] or (0,x), the bug goes away. $\endgroup$ – user1551 Oct 9 '13 at 18:39
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    $\begingroup$ There is special code used to make MathJax play nicely with Markdown in problems and answers. Unfortunately, comments don't use that code, so there are interactions of this kind with the (limited) Markdown available in comments. One horrible kludge to do what you want is to use $\unicode{x5B}a,b)$ to get $\unicode{x5B}a,b)$ when it is followed by a link. $\endgroup$ – Davide Cervone Oct 9 '13 at 20:39
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    $\begingroup$ This issue is the same as was discussed in this older thread, but the behavior of MathJax has since changed so that the solution I gave there is no longer valid. Therefore, I propose that the older question be closed as a duplicate of this one. $\endgroup$ – Zev Chonoles Oct 10 '13 at 3:14
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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 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.

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    $\begingroup$ 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? $\endgroup$ – Zev Chonoles Oct 10 '13 at 3:07
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The mini-markdown in the comments is different from the full-blown markdown in the answer and it's optimized for efficiency even though this means we can't support all the features of normal markdown.

Please adopt a workaround if it is possible.

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