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I couldn't be able to answer this question. A Heyting algebra $(A,\vee,\wedge,\rightarrow,0,1)$ primarily a bounded distributive lattice with top element 1 and bottom element 0. so what about the join of any subset of $A$. Please help..

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    $\begingroup$ I guess any complete boolean algebra can do the job. Take $P(\omega)$ for example? $\endgroup$ – Shervin Sorouri Oct 9 '18 at 15:15
  • $\begingroup$ Of course, exists when we take complete Heyting algebra or complete boolean algebra. But in Heyting algebra may not exist. $\endgroup$ – Kabir Purkit Oct 9 '18 at 15:20
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    $\begingroup$ You asked if infinite join can exist. The answer is yes, as Shervin put it. Bow you say that in some cases, infinite join may not exist (for some subsets), and that is certainly true. But what is you question then? $\endgroup$ – amrsa Oct 9 '18 at 18:14
  • $\begingroup$ @amrsa I asked whether arbitrary join exists or not in a Heyting algebra. $\endgroup$ – Kabir Purkit Oct 10 '18 at 16:45
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Every bounded totally ordered set is a Heyting Algebra.
Not every bounded totally ordered set is a complete lattice, and thus it doesn't have arbitrary joins.
For example, take $C = \{-1/n : n \in \mathbb N\} \cup \{ 1/n : n \in \mathbb N\}$, with the order inherited from $\mathbb Q$.
It is a bounded totally ordered set, with least element $-1$ and greatest element $1$, whence a Heyting Algebra.
However, the set $\{x \in C : x < 0\}$ doesn't have a join.

For more on this topic, see Complete Heyting Algebra.

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