-1

I encountered two similar expressions when prooving two sets A and B are equivalent.

For $A \subseteq B$, we have two ways to prove?

$\forall x \in A$, then $x \in B$

$\forall(x \in A \longrightarrow x \in B)$

  1. Are these two ways both right?

2.how to deal with the situation when A is $\varnothing$ coz in the first way we suppose $A$ is not empty?

Appreciate for your helping hands.

Andrew Ren
  • 171
  • You do not need quantifiers here. Just show $x\in A\implies x\in B$. This is only satisfied if the implication holds for all $x$ and also covers the empty set - case. – Peter Dec 24 '21 at 15:45
  • The statement $\forall x \in A, x \in B$ is just a short hand for $\forall x (x \in A \rightarrow x \in B)$. – Metin Ersin Arıcan Dec 24 '21 at 15:54
  • 1
    "coz in the first way we suppose $A$ is not empty?" No, we don't. It works just fine for empty sets too. – Arthur Dec 24 '21 at 15:57
  • To see that it works fine for $A = \varnothing$ see https://math.stackexchange.com/a/48202/442 – GEdgar Dec 24 '21 at 16:16

2 Answers2

2

Both conditions mean that $\forall x(x\in A\Rightarrow x\in B)$. Forall and implication belong together im math. formulations.

Wuestenfux
  • 20,964
0

Actually there is no problem with the empty set; this is because $p\implies q$ is true whenever $p$ is false, so if $A=\varnothing$ then $x\in A$ is false and because of what I said above we have that the empty set is a subset of every set!

Vivaan Daga
  • 5,531
  • $\forall x \in$ A as a statement of $x$, how can we know its true value. If I plug into a value say 3, $\forall 3 \in$ A, is it true? – Andrew Ren Dec 28 '21 at 03:41