Why is “or” (in logic) sometimes equivalent to “and” (in natural language)?

Consider: "All fruits and vegetables are nutritious"

Rather than:

$$\forall x ((F(x) \land V(x)) \rightarrow N(x)) \text{ Wrong!}$$

it translates as:

$$\forall x ((F(x) \lor V(x)) \rightarrow N(x))$$

But it also translates as (and is indeed equivalent to):

$$\forall x (F(x)\rightarrow N(x))\land \forall x (V(x) \rightarrow N(x))$$

So now we see that "Fruits and vegetables are nutritious" is really just shorthand for "fruits are nutritious and vegetables are nutritious"

Applied to your case:

"the closure of A is the set A and its limit points."

can be translated as:

$$\forall x ((x \in A \lor x \in \text{Lim} A) \rightarrow x \in Closure(A))$$

or as:

$$\forall x (x \in A \rightarrow x \in Closure(A)) \land \forall x (x \in \text{Lim} A) \rightarrow x \in Closure(A))$$

In other words, the confusion is because of the difference between the disjunction of conditions and conjunction of conditionals.

This is a vague question, of course, but I want to try to give some perspective on it based on some thinking I’ve done lately.

I think this is a matter of implicitly using active / passive, which behave dually in a somewhat precise mathematical manner. First, closely observe how we use active and passive regarding the relation of containment:

Let $X$ be a given set, and let $A$, $B$ and $T$ be subsets of $X$. Then we say

• $T$ contains $A$ and $B$, when we mean $T \supseteq A ∪ B$, and
• $T$ is contained by $A$ and $B$, when we mean $T ⊆ A ∩ B$.

This is because we actually mean to avoid repeating some clauses by using the conjunction “and”, in order to shorten sentences. The statements above are (most often) linguistically perceived to be equivalent to respectively

• $T$ contains $A$ and $T$ contains $B$.
• $T$ is contained by $A$ and $T$ is contained by $B$.

So, the conjunction “and” implicitly serves as a conjunction of sentences or propositions.

Note: I don’t mean to define mathematical parlance here, I try to observe it.

Now, if we say that “the closure of $A$ is $A$ and its limits points”, I suspect we immediately understand this to say “the closure of $A$ is the set $T$ containing (exactly) $A$ and all limit points of $A$”. By the above observation, that is the set $T$ such that $T \supseteq A ∪ B$ (when $B$ denotes the limit points of $A$) and $T$ contains nothing more (so $T = A ∪ B$).

However, if we use “and” in other contexts as “$T$ is in $A$ and $B$”, we mean “$T$ is contained in $A$ and $B$”, so $T ⊆ A ∩ B$.

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Mathematically, this can be interpreted e.g. lattice-theoretically: Let $L = (L, ≤, ∨, ∧)$ be a lattice, $∨$ denoting the join (maximum) of two elements, $∧$ denoting the meet (minimum) of two elements. Then there’s a dual lattice $L^{\mathrm{op}} = (L, ≥, ∧, ∨)$ and $L$ and $L^{\mathrm{op}}$ are antitonely isomorphic as lattices.

In the dual lattice $L^\mathrm{op}$, let’s write for all $x, y ∈ L$

• “$y ≤'x$” if $y ≥ x$ in $L$,
• “$x ∨' y$” for $x ∧ y$ in $L$,
• “$x ∧' y$” for $x ∨ y$ in $L$.

So $L^\mathrm{op}$ has order $≤'$, join $∨'$ and meet $∧'$.

If we introduce parlance for the relation, the antitone isomorphism is reflected in our language. For elements $x, y ∈ L$, let’s say $x$ kills $y$ whenever $x ≤ y$ in $L$. (I like killing here because it's so violently vivid.) As humans, we now also automatically say $y$ is killed by $x$ for $x ≤ y$, that is for $y ≤' x$.

Next, observe that for $a, b, t ∈ L$ $$t ≤ a ∧ b \Longleftrightarrow t ≤ a~\text{and}~t ≤ b.$$ Using our parlance: $$\text{$t$ kills $a ∧ b$} \Longleftrightarrow \text{$t$ kills $a$ and $b$}$$ This justifies $a ∧ b$ to be called “$a$ and $b$”.

And the analogous is true for $L^{\mathrm{op}}$ (where we have to add primes “$'$” to all symbols and replace “kills” by “is killed by”). Therefore, for $a, b, t ∈ L$, we have \begin{align*} \text{$t$ is killed by $a$ and $b$} &⇔ t ≤' a ∧' b' \\ &⇔t ≥ a ∨ b \\ &⇔\text{$t$ is killed by $a ∨ b$}, \end{align*} also justiying $a ∨ b$ to be called “$a$ and $b$”.

The difference lies in which perspective we take on an implicit relation – from above or from below. And it depends on whether we use (implicit) verbs describing the relation in active or passive.

In our case, $L$ is the powerset of some space $X$ with $\supseteq$ as order and killing is containing.

This also explains why such a confusion doesn't happen when lingustically using “or”: We never say “$A$ or $B$” as in “the closure of $A$ is $A$ or its limit points”. This is because $t ≤ a ∨ b ⇔ \text{$t ≤ a$ or $t ≤ b$}$ does not hold for general lattices (think of disjoint union), so there’s no justification for saying “$a$ or $b$” for $a ∨ b$; and there need not be any other element fulfilling this universal property: There is (in general) no set “containing exactly $A$ or its limit points”.

(And in a sense, this generalises to categories with limits and colimits.)

It depends on how you word it. Does $A \cup B$ correspond to "or" (as the notation $\vee$ for "or" suggests), or does it correspond so "and"?

You argue that $A \cup B$ corresponds to "and", since $$ A \cup B\quad\text{consists of both}\quad A\quad\text{and}\quad B $$

The reasoning I guess used by Boole and his followers is: $A \cup B$ corresponds to "or" since $$ A \cup B = \{x\;:\; x \in A \text{ or } x \in B\} $$