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In the above lattice is the GLB of b and c unique? I think the GLB of b and c should be only d. In my book it's given that the GLB of b and c is not unique as GLB of b and c can even be a.

I think that GLB of b and c would not have been unique if d and a would have been at same level which is not the case here.

Can someone tell me whether I am right?

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By definition, the GLB (or $\inf$, for infimum), is the greatest lower bound. Looking at your Hasse diagram, the set $\{b,c\}$ has two lower bounds, namely $a$ and $d$. However, these two (i.e. $a$ and $d$) are not comparable, so there is no maximum among the lower bounds. Hence, the GLB of $b$ and $c$ does not exist. In particular, this tells you that your ordered set is not a lattice.

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  • $\begingroup$ Why a and d are not comparable ? d is at a higher level than a in the hasse diagram right ? So it should be greater. $\endgroup$ – Zephyr Jul 23 '17 at 5:18
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    $\begingroup$ Higher has nothing to do with it. There is no line between them. The same partial order could be diagramed with d below a. $\endgroup$ – William Elliot Jul 23 '17 at 12:02
  • $\begingroup$ If there would have been a line between d and a, then they are not interchangeable right ? And d and e are not interchangeable right, as it would mean eRb rather than bRe which is not correct .Am I correct about the second statement ?@William Elliot $\endgroup$ – Zephyr Jul 23 '17 at 15:06
  • $\begingroup$ @Zephyr As William says, the order relation is indicated by the lines; the height at which the elements are drawn means nothing if there's no line. If there were a line joining $a$ and $d$ and $a$ was lower than $d$, then we would have $a\le d$, but this is not the case. $\endgroup$ – Reveillark Jul 23 '17 at 16:50
  • $\begingroup$ Can you tell me if my second statement about exchanging d and e is correct or not? $\endgroup$ – Zephyr Jul 23 '17 at 16:50

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