I'm not sure of the correct notation if someone could please help.

Say you wish to have a set comprised of the union of two sets, such as $$h=\Big\{ \begin{bmatrix}x & y\\y & x\end{bmatrix}: x^2-y^2=1 \Big\};$$$$g=\Big\{ \begin{bmatrix}x & y\\-y & -x\end{bmatrix}: x^2-y^2=1 \Big\}.$$

I'm not sure if putting the union sign inside the set is correct. Would it be $$\phi = \Big\{ \begin{bmatrix}x & y\\y & x\end{bmatrix}\cup \begin{bmatrix}x & y\\-y & -x\end{bmatrix}\ : x^2-y^2=1 \Big\}?$$

  • $\begingroup$ $\phi=h\cup g$ or replace the $\cup$ in your final line by a comma. $\endgroup$
    – Jonathan
    Commented Feb 15, 2014 at 17:59
  • 1
    $\begingroup$ Thanks, that's all I needed to know :) $\endgroup$
    – user112633
    Commented Feb 15, 2014 at 18:00
  • $\begingroup$ Ignore the example, it's just made up. $\endgroup$
    – user112633
    Commented Feb 15, 2014 at 18:01

2 Answers 2


You can

$(1)$: Provided you've already defined $h, g$, as you did in this post, the simplest route is to express $\phi$ as the union of $h, g$, as in $$\phi = h\cup g$$

$(2)$ Write $$\phi = \left\{ \begin{bmatrix}x & y\\y & x\end{bmatrix}\text{ or }\; \begin{bmatrix}x & y\\-y & -x\end{bmatrix}\ : x^2-y^2=1 \right\}$$

The union operation is an operation on sets; for example, $h, g$ are sets.

Instead of "or", you can use the notation $\lor$ in $(2)$.

Better yet, if you really like set-builder notation, you can write:

$$\phi = \{A: A \in g \lor A \in h\}$$ which means: "The set of all matrices A that belong to $h$ or belong to $g$.

You might also want to include the field to which $x, y$ belong. E.g., $x, y \in \mathbb R$.

  • $\begingroup$ So you wouldn't include set brackets in the case of (1)? $\endgroup$
    – user112633
    Commented Feb 15, 2014 at 18:07
  • $\begingroup$ That's exactly right. Once $g, h$ are defined, we can express $\phi$ as the union of two sets, which is, again, a set. $\endgroup$
    – amWhy
    Commented Feb 15, 2014 at 18:09
  • $\begingroup$ @amWhy: Have you seen that notation in $(2)$ used before? I think I would find it confusing if it weren't in context that made it obvious what was meant. $\endgroup$
    – user14972
    Commented Feb 15, 2014 at 18:10
  • $\begingroup$ @amWhy Thank you, I would upvote, but it won't let me log in to my account; probably because I am at uni. I will do later. $\endgroup$
    – user112633
    Commented Feb 15, 2014 at 18:12
  • $\begingroup$ I would not use $(2)$. If using set-builder notation, I'd use $$\phi = \{A: A \in g \lor A \in h\}$$ which essentially means $\phi = g\cup h$. $\endgroup$
    – amWhy
    Commented Feb 15, 2014 at 18:12

Maybe $$ \left\{\,\begin{bmatrix}x&y\\ex&ey\end{bmatrix}\in \mathbb R^{2\times 2}: x^2-y^2=1, e=\pm1\,\right\}$$ This matches the set-builder syntax $\{\,x\in X: \phi(x)\,\}$ as per Axiom schema of Specification (or Comprehension). Depending on context, the class-builder syntax $\{\,x:\phi(x)\,\}$ may be ok, but at any rate having a "complex expression" to the left of the colon is at least ambiguous as it may suggest a replacement $\{\,f(x):x\in X\,\}$. (Actually, the matrix itseld is already such a "complex expression", so this is arguably an instance of replacement anyway).


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