What does $\wedge$, $\vee$ and $\triangle$ mean in real analysis?

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The context where I found them is the following.

For a measurable space $(X,M)$ and $f, g : X → \mathbb{R}, a \in \mathbb{R}$ then $a · f, f + g$, and $f · g$ are all measurable. Moreover, if $f, g : X → \mathbb{R}$ then $f ∧ g$ and $f ∨ g$ are measurable, as is $|f|$.

and

$57$ (metric on $L_C(M)$) Let $(X,M, µ)$ be a finite metric space (ie. $µ(X) < ∞$). Define $d$ on $L_C(M)$ via $d(f, g) = \int |f − g| ∧ 1dµ = d(f − g, 0)$.

for $\triangle$ (in the last sentence),

$62$ (Approximation properties of $m^n$ ) We let $m^n$ be the completion of $m × · · · × m$ where $m$ is the Lebesgue measure on $\mathbb{R}$. So $L^ n$ is the Lebesgue measurable sets on $\mathbb{R}^ n$ . Take $E ∈ L^n$ . Then

  1. $m^n (E) = \inf\{m^n (O) | E ⊆ O$ and $O$ is open$\} = \sup\{m^n (K) | K ⊆ E$ and $K$ is compact$\}$.
  2. $E = A_1\setminus N_1$ where $A_1$ is $G_δ$ and $m^n (N_1) = 0$, $E = A_2 ∪ N_2$ where $A_2$ is $F_σ$ and $m^n (N_2) = 0$
  3. $m^n (E) < ∞$ implies $∀\epsilon > 0, ∃(R_j ) ^N _{j=1}$ of disjoint open rectangles such that $m^n (E \triangle (∪R_j )) = 0$

Seems $\wedge$ and $\vee$ are used between functions, and it looks the wedge product of differential forms in the second paragraph, not sure if it is since this is the real analysis class and we didn't cover any of differential geometry.

And $\triangle$ seems to be used between sets and I think it is the same as the set minus operator '$\setminus$', I don't know why the $\epsilon$ comes up and guess it is related to $\triangle$ somehow?

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  • $\wedge$ and $\vee$ are $\min$ and $\max$, respectively. (See, e.g., this.)

  • $\Delta$ is the symmetric difference: $A\Delta B = (A\setminus B)\cup (B\setminus A)$.