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My teacher claims that when an equation in variables $x_1,x_2,\ldots,x_n$ has no solutions, you should denote this fact with $(x_1,x_2,\ldots,x_n)\in\varnothing$.

An empty set can't have an element in it, so this can't be right.

user26486
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    Your teacher's idiosyncratic notation means exactly what is desired, namely that there are no elements in the set of all solutions. – vadim123 Nov 20 '14 at 19:43
  • To be an element of an empty set is rather strange. We just had to write down $\varnothing$ for no solution but we were not allowed to write the {} symbols around it. – imranfat Nov 20 '14 at 19:44
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    @imranfat The sets $\varnothing$ and ${\varnothing}$ are different. One of them has an element, and the other doesn't. – Arthur Nov 20 '14 at 19:45
  • "should denote"? I'd say "can denote". It's not wrong, but it's not notation that is used for this very often. – Simon S Nov 20 '14 at 19:46
  • $(x_1,x_2,\ldots,x_n)\in\emptyset$ is equivalent to the original equation. – Minimus Heximus Nov 20 '14 at 19:50
  • The above words 'can' and 'idiosyncratic' are apropos here. – copper.hat Nov 20 '14 at 19:52
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    An equation (or a system of equations) in so many variables has a solution set $S$ which is a subset of the universe $\Omega$ for which the equations make sense. When there are no solutions we write $S=\emptyset$. A statement as $(x_1,x_2,\ldots, x_n)\in\emptyset$ seems pretty forlorn to me. – Christian Blatter Nov 20 '14 at 20:22
  • I think what the teacher actually means is ${(x_1, \dots, x_n)} = \varnothing$ (which is still a slight and reasonable abuse of notation, but much better than the original). – anomaly Nov 20 '14 at 20:35
  • @anomaly Could you explain why you think it is any better? Both notations make it seem like $(x_1,\ldots,x_n)$ can be a member of an empty set and yours is not any better for me, maybe even worse. You're claiming that (let the solution be $x$) ${x}={}$ is a better notation, while $x\in\mathbb {}$ is worse. The only difference here is that $\varnothing$ could contain other variables other than $x$ in my teacher's example, while in your example $\varnothing$ only contains $x$. But it doesn't change the fact that the non-existent variable $x$ is there. – user26486 Jan 22 '15 at 17:24
  • @anomaly You imply that it is fine for $\varnothing$ to contain a non-existent variable as long as it doesn't contain anything else. Why do you find your notation better? – user26486 Jan 22 '15 at 17:29
  • It's an abbreviation for $\left{(x_1, \dots, x_n):, \text{some condition on the $x_i$ holds}\right} = \varnothing$, which is unobjectionable. It's not a great abbreviation (and not one that I'd use myself), but the meaning is clear from context. – anomaly Jan 22 '15 at 17:37
  • @anomaly It is just your notation for the set of all solutions. As some answers here showed, we can create non-controversial notation for that, like ${(x_1,\ldots,x_n):f((x_1,\ldots,x_n))=0}=\varnothing$, where $f((x_1,\ldots,x_n))=0$ is an equation equivalent to the one solved. – user26486 Jan 22 '15 at 17:57

4 Answers4

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You can form the set $L$ of all solutions. So iff $(x_1, \cdots, x_n)$ is a solution, then $(x_1, \cdots, x_n)\in L$. Now, if there are no solutions, then $L = \varnothing$. Insert that, and you get $(x_1, \cdots, x_n)\in \varnothing$.

In other words, the sequence $(x_1, \cdots, x_n)$ is a solution to an impossible set of equations iff $(x_1, \cdots, x_n)\in \varnothing$.

However, saying that "$(x_1, \cdots, x_n)\in \varnothing$ means there are no solutions" is taking it a bit far. I would rather say "$L = \varnothing$ means there are no solutions".

Arthur
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You are correct that the empty set is the set with no members. The solution set of an equation is, however defined as the set of all solutions that satisfy the equation and since that set has no members the solution set is empty.

Thus, saying that the equation has no solution, and saying that the solution set of the equation is empty mean the same thing.

Recall that the solution set is the set of all values that turn an open sentence into a true statement.

Thus, the membership symbol is an abbreviation for the fact that the variable can be replaced by a value that is "in", "belongs to", or "is a member of" the set. If the set is empty, then the variable can not be replaced by any value. This is what it means for an equation to have "no solution", there is no value that when substituted for the variable will result in a true statement.

I agree with you that the notation does lead to some confusion when dealing with the special case of the solution set being empty. In this case "what the variable can be replaced by belongs to" a set with no members. But, if you like, you can simply say "No solution." Note that the meaning here is still the value that the variable can be replaced by belongs to the replacement set, which happens to be empty so there is no value.

variable: A symbol used to represent any member of a given set.

domain of a variable: The set whose members may serve as replacements for the variable; also called replacement set.

cybersecurity attacker
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  • Of course, surely the set of all solutions is empty in this case, but how can an object exist in a fully empty set? If you think this is correct notation, then should $A\in\varnothing$, where $A={A}$ also be correct? In both cases, the object is impossible and if you agree with my teacher's claim, then you agree with this claim, thus overall agreeing that all impossible objects exist in the empty set. – user26486 Nov 20 '14 at 21:31
  • Again, yes, the solution set is empty. I don't know how you defined an object, but let it be something that can exist in a set (which is just a collection). $(x_1,\ldots,x_n)$ is an object and thus should be in a set. But $(x_1,\ldots,x_n)\in\varnothing$ implies it is not able to be a member of a set and therefore it is not a possible object. $(x_1,\ldots,x_n)\in\varnothing$ thus implies an impossible object is in the empty set. But that means that all impossible objects exist in the empty set. You agree with my teacher's claim iff you agree with this claim, unless I'm missing something. – user26486 Nov 20 '14 at 21:52
  • It simply means $\not\exists(x_1,\ldots,x_n)(f(x_1,\ldots,x_n)=0)$, where $f(x_1,\ldots,x_n)=0$ is the equation we're trying to solve. In your case, $f(x)=x+2-x-1=1$ and so it is true that $\not\exists x(f(x)=0)$, since $\forall x\in\mathbb R, f(x)=1$. – user26486 Nov 20 '14 at 22:04
  • $x$ is an object. $\varnothing$, or ${}$, is a set with no objects. $\in$ means 'is in'. Therefore, $x\in\varnothing$ denotes '$x$ is an object in a set with no objects'. – user26486 Nov 20 '14 at 22:17
  • If you will, let us call it 'something' instead. $x$ is something. ${}$ contains nothing. How can something be where nothing is? – user26486 Nov 20 '14 at 22:23
  • But if you denote it as $x$, then it is automatically something - a variable always denotes something, whether it be a number or a vector or etc. And yet the empty set contains nothing, so it can't contain a variable. As for your $x=\frac{0}{0}$, any mathematician would agree that there exist no solutions to this equations (while you're claiming $x=\frac{0}{0}, \forall x\in\mathbb R$). – user26486 Jan 22 '15 at 17:14
  • It is completely correct to denote $S=\varnothing$ or $S\subseteq\varnothing$, where $S$ is the set of all solutions, but, as Wikipedia says, a variable is a mathematical object, and you have to realize that a mathematical object cannot exist in an empty set which contains nothing. Though you can use it in a proof by contradiction, as in "but $x\in\varnothing$. Therefore, we have a contradiction and $\not\exists x$" When we solve equations with no solutions, we implicitly prove it by contradiction - we assume the mathematical object $x$ exists, but we get it cannot exist, thus $\not\exists x$. – user26486 Jan 22 '15 at 17:49
  • I think you're not thinking about this rigorously. That it "represents" the empty set is true in some sense - it can contain no value, just like the empty set can contain no member. But the membership symbol rigorously means that $x\in S\implies x$ is a member of $S$, but by definition $\varnothing$ has no members. – user26486 Jan 22 '15 at 18:03
  • No, a variable can't represent two or more numbers. Don't mix the definition of a variable with the definition of a set - a variable is one explicit number (or a vector or another object). When we say $x^2+3x+2=0\implies x\in\mathbb {1,2}$, then what we're saying is that if it happens so that $x^2+3x+2=0$, then we know that $x$ is either $2$ or $3$. – user26486 Jan 22 '15 at 18:06
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I would write $card\{X \mid f(X) = 0\} = 0 $ where $card\{S\}$ is the number of elements in a set $S$ and $X$ is a vector $(x_1, x_2, ..., x_n)$.

Another way would be $\{X \mid f(X) = 0\} =\varnothing $.

marty cohen
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A more efficient way to denote that the equation has no solution, is by a Boolean "false", "F", or "$\bot$", consistent with your observation that the empty set cannot have any elements.

Jan Stout
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