Isn't $k\in \emptyset$ a contradictory statement?

Question

I've seen $k\in \emptyset$ written in the context of empty sum. I read, $k\in \emptyset$ as "$k$ is an element of the set which has no elements." To me, this sounds like a contradiction. If the empty set contains the element $k$ it is no longer empty.

Also, how do you arrive at the conclusion that because $k\in \emptyset$, $k$ does not exist? To say that $k$ is in the $\emptyset$ is a false statement, therefore nothing follows from it. If we want to say that $k$ does not exist we simply write $\neg \exists k$.

Since $k\in \emptyset$ is widely used there must be an explanation. Can you clarify how you rationalize the contradiction that empty set is not really empty? Does the set theory has its own logic where contradiction is legal?

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Edit

I'm sorry it was not clear from the question that $k$ is a summation index and as such it cannot be a free variable. $k$ takes natural numbers as its values $k\in \mathbb{N}$.

I think, therefore, answers claiming that $k\in \emptyset$ cannot be false because $k$ is a free variable are not acceptable. Please correct if I misunderstand.

Answer 1

The formula $$k\in\emptyset \tag1$$ is a predicate with a free variable $k$. It is still not true nor false, because you still have not said what $k$ is. Please note that this predicate is not saying that there exists $k$ with $k\in\emptyset$. That formula would be $$\exists k,\ \ k\in\emptyset \tag 2$$ which indeed has the value false. This is probably what you have in mind.

Again, the predicate $(1)$ is a formula with a free variable, so it doesn't have a truth value, while the proposition $(2)$ has the value false. On the other hand, what you observe is that there cannot be a $k$ for which formula $(1)$ has the value truth. This sentence is captured by the formula $$\forall k, \ \ \neg(k\in \emptyset). \tag 3$$ which has the value true.

Answer 2

You have basically two seperate questions:

On the notation for summations and specifically the empty sum, we read:

$$\sum_{k \in X}a_k$$ As the sum of all $a_k$ such that $k \in X$. For the empty sum, we would read: $$\sum_{k \in \emptyset}a_k$$ As the sum of all $a_k$ such that $k \in \emptyset$, which (you are correct) there are none. Which is why the empty sum evaluates to $0$, because we are summing nothing. This is simply a matter of notation and how to read summations, it is not a logical argument, no worries!

On your question about contradictions you write two things:

How do you arrive at the conclusion that because $k \in \emptyset$, $k$ does not exist? To say that $k$ is in the $\emptyset$ is a false statement, therefore nothing follows from it.

Your second statement is incorrect. If we can come to the conclusion that a false statement, namely $k \in \emptyset$, is true, then we can come to the conclusion that any other false statement is also true, which we know is not correct. Therefore, we know our original statement was incorrect, and hence there are no $k \in \emptyset$. This is the concept known as the principle of explosion and it relies around the idea that (If $F$ is a false statement). Using this concept, when we come to a statement that we know is false, we can then deduce that our original assumption was false. For example, if we assume $A$ to be true then:

$$A \Rightarrow B \Rightarrow C \Rightarrow F$$ Would point us to the idea that $A$ is false, since we know that $F$ is false and when we assume $A$ to be true $F$ evaluates to true. In proofs we state that this is contradiction, since $F$ should be true based on our work but we know it is false.

To arrive at the conclusion (with a contradiction) that $(k \in \emptyset) \Rightarrow \neg \exists k$, we can do the following: First, we assume that there exists a $k$ such that $k \in \emptyset$. Then, since $\emptyset \subset A$ for any set $A$, $k \in A$. But the set $A = \{b\}$ does not contain $k$, hence this is a contradiction and thus there does not exist a $k$ such that $k \in \emptyset$. Or any other argument where you can arrive at a contradiction that is shorter or longer than mine. A much easier proof, is just to say: by definition $\neg \exists k (k \in\emptyset)$.

Answer 3

What is a contradiction in Mathematical Logic?

Definition: A sentence is called a contradiction if its truth table only has false values.

Where a sentence is a formula with no free variables.

The formula x$\in$$\emptyset$ needs to unraveled, as $\emptyset$ is a defined term.

formally, x$\in$$\emptyset$ is shorthand for x$\in$y $\wedge$ $\forall$w$\in$y(w ≠ w)

Note: The same unraveling needs to be done with the empty sum in order to assess if it is a contradiction or not. The standard notation is just a shorhand for a longer formula, don't focuse on the syntax for the shorthand.

So x, is a free variable ( not bound by a quantifier) and so it doesn't make sense to talk about if it's a contradiction because it isn't even a sentence.

If we wanted to consider the sentence:

$\exists$x$\exists$y(x$\in$y $\wedge$ $\forall$w$\in$y(w ≠ w))

Then we could say that such a sentence is a contradiction

I usually see x$\in$$\emptyset$ as a premise in a vacuously true implication.

For example, Let's Prove $\emptyset$ is a subset of any set.

Definition: X $\subset$ Y $\iff$ $\forall$z(z$\in$X $\rightarrow$ z$\in$Y)

So we prove $\forall$z(z$\in$$\emptyset$ $\rightarrow$ z$\in$Y)

Proof: Let Y be a set.

Let $z_0$ be arbitrarily chosen.

Then

$z_0$$\in$$\emptyset$ $\rightarrow$$z_0$$\in$Y

Is true as the premise is false.

Thus $\emptyset$ is a subset of every set.