# Defining arbitrary join on the set of complete ideals of a Heyting Algebra

Given a Heyting Algebra $$H$$ we define a complete ideal (or c-ideal) $$I$$ to be a subset of $$H$$ satisfying.

1. $$\bot \in I$$
2. $$b \in I$$ and $$a \leq b$$ implies $$a \in I$$
3. $$X \subseteq I$$ and $$\bigvee X$$ exists in $$H$$ implies $$\bigvee X \in I$$.

Note $$H$$ is not necessarily complete so arbitrary joins do not neccesarily exist in $$H$$. A complete ideal generalizes the notion of an ideal from binary joins to arbitrary joins if they exist.

We can use c-ideals to complete the Heyting Algebra $$H$$. Let $$H'$$ be the set of c-ideals of $$H$$. Let $$I_{\alpha}$$ be a family of c-ideals where $$\alpha \in \mathcal{J}$$ some index set.

We define the meet in $$H'$$ to be the intersections of $$I_{\alpha}$$ that is $$\underset{\alpha \in \mathcal{J}}{\bigwedge} I_{\alpha} = \underset{\alpha \in \mathcal{J}}{\bigcap} I_{\alpha}$$. It is easy enough to show that the intersection of c-ideals is a c-ideal and it is in fact the meet. Unfortunately, the union of c-ideals is not, in general, a c-ideal. So for the join things are a bit more complicated. We define the join below

$$\underset{\alpha \in \mathcal{J}}{\bigvee} I_{\alpha} = \{ \bigvee X \ | \ X \subseteq \underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha} \ \text{and} \ \bigvee X \ \text{exists in H} \}.$$

I am struggling to show that this object is even a c-ideal, in particular on showing 2. and 3. hold.

Attempt:

1. $$\emptyset \subseteq \underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha}$$ and $$\bigvee \emptyset = \bot$$ exists in $$H$$ so $$\bot \in \underset{\alpha \in \mathcal{J}}{\bigvee} I_{\alpha}$$
2. Let $$b \in \underset{\alpha \in \mathcal{J}}{\bigvee} I_{\alpha}$$ and $$a \leq b$$. There exists $$X \subseteq \underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha}$$ such that $$\bigvee X = b$$. We need to show that there exists $$Y \subseteq \underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha}$$ such that $$\bigvee Y = a$$. Some candidates for $$Y$$ are $$\downarrow a$$, $$X \cap \downarrow a$$ and $$\underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha} \cap \downarrow a$$. The upside of $$\downarrow a$$ is that $$\bigvee \downarrow a = a$$ but there is no way to show $$\downarrow a \subseteq \underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha}$$. The upside of the latter two is that they obviously satisfy the subset requirement but there is no obvious way to show that there join is $$a$$.
3. Let $$Y \subseteq \underset{\alpha \in \mathcal{J}}{\bigvee} I_{\alpha}$$ and $$\bigvee Y$$ exist in $$H$$. To show $$\bigvee Y \in \underset{\alpha \in \mathcal{J}}{\bigvee} I_{\alpha}$$ it suffices to show $$Y \subseteq \underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha}$$. Let $$x \in Y$$ then by assumption $$x \in \underset{\alpha \in \mathcal{J}}{\bigvee} I_{\alpha}$$. Thus for some $$X \subseteq \underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha}$$ we have $$x = \bigvee X$$. Again I am stuck because we cannot say $$x \in X$$ and conclude $$x \in \underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha}$$.

Reference: [1] A.S. Troelstra, D. Van Dalen. Constructivism in Mathematics An Introduction vol. 2.

Here is some insight for you.

Consider a partially ordered set $$S$$ which has all meets. We will prove that $$S$$ has all joins.

Let us consider some indexed collection $$\{s_i \in S\}_{i \in I}$$. Let $$J = \{x \in S \mid \forall i \in I (s_i \leq x)\}$$. I claim that $$\bigwedge J$$ is the join of $$s$$.

Indeed, we first note that for all $$i$$, for all $$x \in J$$, $$s_i \leq x$$ (by the very definition of $$J$$). Therefore, $$s_i \leq \bigwedge J$$.

Now suppose that for all $$i \in I$$, $$s_i \leq x$$. Then $$x \in J$$, so $$\bigwedge J \leq x$$. This completes the proof. $$\square$$

Your definition also works. Let's work through it.

First, we must show that $$Q := \{ \bigvee X \mid X \subseteq \bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha$$ and $$\bigvee X$$ exists$$\}$$ is in fact a $$c$$-ideal.

As you have shown, (1) is trivial.

For (2), suppose that $$X \subseteq \bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha$$ and $$\bigvee X$$ exists, and that $$a \leq \bigvee X$$. Then $$a = a \land \bigvee X = \bigvee \{a \land x \mid x \in X\}$$ (using the fact that $$a \land$$ is the left adjoint to $$a \implies$$ and therefore preserves any colimits that do exist). Since each $$I_\alpha$$ is downward closed, if $$x \in I_\alpha$$ then $$a \land x \in I_\alpha$$. And thus, we see that $$\{a \land x \mid x \in X\} \subseteq \bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha$$. Thus, $$a \in Q$$.

For (3), we need to get a big cleverer if we wish to avoid choice. Given $$a \in Q$$, let $$f(a) = \{w \in \bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha \mid w \leq a\}$$; then clearly, $$a = \bigvee f(a)$$.

Now suppose we have some family $$\{x_k \in Q\}_{k \in K}$$, and that $$\bigvee\limits_{k \in K} x_k$$ exists. Then in particular, we have $$\bigvee\limits_{k \in K} x_k = \bigvee\limits_{k \in K} \bigvee f(x_k) = \bigvee \bigcup \limits_{k \in K} f(x_k)$$. Note that $$f(x_k) \subseteq \bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha$$ by definition, so $$\bigcup \limits_{k \in K} f(x_k) \subseteq \bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha$$. Therefore, $$\bigvee\limits_{k \in K} x_k = \bigvee \bigcup \limits_{k \in K} f(x_k) \in Q$$.

Now, let's go about proving that $$Q$$ is in fact the join of the $$I_\alpha$$.

Clearly, we see that if $$x \in I_\alpha$$, then $$x = \bigvee \{x\} \in Q$$. So $$I_\alpha \subseteq Q$$ for all $$\alpha$$.

Now suppose that we had some $$y$$ such that for all $$\alpha$$, $$I_\alpha \subseteq y$$. Then condition 3 ensures that $$Q \subseteq y$$.

• Are you familiar with the troelstra reference? Is there any reason why he didn’t just use this well known property that you outlined in the first part of your answer? Commented Dec 27, 2021 at 22:50
• @ToucanIan I haven't read it. I suspect that the explicit form of the join will be important when showing that the result is a Heyting Algebra (since it suffices to show that $a \land \bigvee\limits_{i \in I} x_i = \bigvee\limits_{i \in I} a \land x_i$ to show that $a \land$ is a left adjoint). Commented Dec 27, 2021 at 22:52
• This was my suspicion. Thanks! Commented Dec 27, 2021 at 23:37
• Can you say more about your argument in 2. involving left adjointness of $\land$. Is there a more elementary route to this? Commented Dec 27, 2021 at 23:51
• @ToucanIan We have $a \land \bigvee X \leq y$ iff $\bigvee X \leq a \Rightarrow y$ iff $\forall x \in X (x \leq a \Rightarrow y)$ iff $\forall x \in X (a \land x \leq y)$ for any $y$, thus proving that $a \land \bigvee X$ is $\bigvee \{a \land x \mid x \in X\}$. Note that we use the $\land-\Rightarrow$ adjunction. Commented Dec 27, 2021 at 23:58

I found an alternative way to define the join. I would still like insight on the original question and/or a relationship to what follows.

Let $$A = \{ K \in H' \ | \ \underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha} \subseteq K \}$$ that is $$A$$ is the set of complete ideals containing the union. Now clearly, $$A$$ is inhabited since $$\underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha} \subseteq H \in H'$$. Trivially we have the $$\bigcap A$$ is a complete ideal as it is an intersection of complete ideals. We wish to show that $$\underset{\alpha \in \mathcal{J}}{\bigvee} I_{\alpha} = \bigcap A$$. First note that for every $$\beta \in \mathcal{J}$$ we have $$I_{\beta} \subseteq \underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha} \subseteq \bigcap A$$ (where the last inclusion comes from the fact that $$\underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha} \subseteq K$$ for all $$K \in A$$.) So clearly, $$\bigcap A$$ is an upperbound. Suppose you have another upperbound $$M$$ in $$H'$$. That is for all $$\beta \in \mathcal{J}$$ we have $$I_{\beta} \subseteq M$$. Then $$\underset{\alpha \in \mathcal{J}}{\bigcup} I_{\alpha} \subseteq M$$ and thus $$M \in A$$. So we can conclude that $$\bigcap A \subseteq M$$.

• Your alternate proof is essentially the same idea as the proof in my answer that any partial order with all meets has all joins (except that you have to replace $\bigcup\limits_{\alpha \in \mathcal{J}} I_\alpha \subseteq K$ with $\forall \alpha \in \mathcal{J} (I_\alpha \leq K)$, since you don't have an external notion of joins to rely on). Commented Dec 27, 2021 at 22:40
• I’m confused by your comment. I’m on saying $K$ contains the union of the family of ideals. I’m not mentioning any joins. Commented Dec 27, 2021 at 23:36
• The union is the "external notion of join". You're trying to take a subset of the complete Heyting algebra $P(H)$ and prove it's a complete Heyting algebra, and you're relying on the join in the larger algebra $P(H)$ - that is, the union. Commented Dec 28, 2021 at 0:02
• Doesn’t then $Q$ in your original answer rely on that same external join? Commented Dec 28, 2021 at 0:04
• Yes, it does. But I was referring to the first proof I gave at the top of the answer that any poset with all meets has all joins, which does not use $Q$ at all and ends with a $\square$. Commented Dec 28, 2021 at 0:05