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My linear algebra instructor claims that the expression for a set of vectors

$$ B = \left\{ \vec v \in \mathbb{R^2} : \vec v = a \begin{bmatrix}2 \\ 1\end{bmatrix} \right\}, \text{ where } a \in \mathbb{R} \tag{1}$$

is incorrect, whereas

$$ B = \left\{ \vec v \in \mathbb{R^2} : \vec v = a \begin{bmatrix}2 \\ 1\end{bmatrix}, \text{ where } a \in \mathbb{R} \right\} \tag{2}$$

is correct.

His reasoning is that the variable $a$ doesn't exist anymore once the set is closed, so there is nothing to be quantified, and the first expression is meaningless.

On the other hand, a definition in the textbook we are using indicates that an expression such as

$$ \text{Let } a \in \mathbb{R}. \text{ Then, } B = \left\{ \vec v \in \mathbb{R^2} : \vec v = a \begin{bmatrix}2 \\ 1\end{bmatrix}\right\} \tag{3}$$

is correct.

I've also learned in another class, which tackled formal logic a little more deeply than this linear algebra class, that quantifying a variable before or after a statement involving the variable is mostly an arbitrary/stylistic decision on the part of the writer. If we can quantify a variable before a statement, we can quantify it after a statement as well. According to this other class, the first statement might be okay, although I don't remember tackling sets specifically when working with quantifiers.

I know there is such a thing as quantifier scope in formal logic, but I'm not sure whether set brackets delineate quantifier scope in mathematical statements such as the above, as my linear algebra instructor seems to suggest.

Since ultimately a statement such as this should make sense when expressed in natural language, I think the first statement is one where the speaker mentions, after specifying the set, that, "by the way, the $a$ in the preceding statement (which happens to contain a set in this case) is a real number, in case that wasn't clear", which is similar to how a speaker might mention the domain of the variable prior to specifying the set, as in statement $(3)$.

I was wondering whether anyone can confirm or deny my instructor's claim, and whether the first statement is truly a mathematically meaningless statement.

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    $\begingroup$ Both (1) and (2) are meaningful, but they are not the same. The set in (1) has exactly one element, which element depends on the pregiven value $a$, while the set in (2) has as many members as many value $a$ can take from $\Bbb R$. $\endgroup$
    – Berci
    Sep 18, 2019 at 17:54
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    $\begingroup$ But "for some $a\in\mathbb{R}$", inside the set definition, means that the set does have an element for each possible value of $a$, since all those possibilities would make the existential statement true. $\endgroup$ Sep 18, 2019 at 18:14
  • $\begingroup$ @Berci Since there is no pregiven value of a, that would make (1) meaningless, no? $\endgroup$
    – Chrisuu
    Sep 18, 2019 at 21:59
  • $\begingroup$ But I think I'm starting to see what you mean. It doesn't make sense to quantify the variable prior to the set, because then the set would only contain one element. And quantifying it after the set is similar to quantifying it before. Maybe the variable does have to be quantified within the set after all... And I guess that makes $(3)$ distinct from $(2)$ as well. Regardless what we pick for $a$, the set in $(3)$ will contain just one element, namely the vector that's been scaled by $a$. $\endgroup$
    – Chrisuu
    Sep 18, 2019 at 22:04

1 Answer 1

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  1. The set $$ B = \left\{ \vec v \in \mathbb{R^2} : \vec v = a \begin{bmatrix}2 \\ 1\end{bmatrix}, \text{ where } a \in \mathbb{R} \right\} \tag{2},$$ that is, $$ B = \left\{a \begin{bmatrix}2 \\ 1\end{bmatrix}:a \in \mathbb{R} \right\} \tag{2},$$ is the set of vectors of the form $a\begin{bmatrix}2 \\ 1\end{bmatrix},$ where $a$ is real; in other words, each element of $B,$ for some real $a,$ equals $a\begin{bmatrix}2 \\ 1\end{bmatrix}.$ This is the set that you want.
  2. On the other hand, the statement $$ B = \left\{ \vec v \in \mathbb{R^2} : \vec v = a \begin{bmatrix}2 \\ 1\end{bmatrix} \right\}, \text{ where } a \in \mathbb{R} \tag{1}$$ means that for some (pre-specified) real $a,$ the set $B$ contains precisely the element $a \begin{bmatrix}2 \\ 1\end{bmatrix}.$
  3. Similarly, the statement $$ \text{Let } a \in \mathbb{R}; \text{ then, } B = \left\{ \vec v \in \mathbb{R^2} : \vec v = a \begin{bmatrix}2 \\ 1\end{bmatrix}\right\} \tag{3}$$ says that $B$ is a single-element set, but this time $a$ is an arbitrary, instead of a pre-specified, real number.
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