Quantifying a variable in a set after specifying the set

135 Views Asked by At

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.

1

There are 1 best solutions below

0
On
  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.