Confusion in understanding why empty set is a subset of every set

Question

statement 1 : If $x \in \emptyset$ then $x \in A$.

statement 2 : If $x \in \emptyset$ then $x \notin A$.

I know that both statements are true since the hypothesis is false.

But first statement says that $\emptyset \subset A$ while second statement says that $\emptyset$ is not a subset of $A$.

My question is why we prefer $\emptyset \subset A$ over the other implication that $\emptyset$ is not a subset of $A$? Thanks.

"$x \in \emptyset$ then $x \notin A$ (..) says that $\emptyset$ is not a subset of $A$". .. if such an $x$ exists, which does not. — mvw, Jul 27 '17 at 11:30

score 6 · Answer 1 · answered Jul 27 '17 at 11:13

6

The statement

(For all $x$,) if $x\in B$ then $x\notin A$.

does not say that $B$ is not a subset of $A$. In order to say that $B$ is not a subset of A, you would need to say

There is an $x\in B$ such that $x\notin A$.

This is vacuously false then $B$ is $\varnothing$.

answered Jul 27 '17 at 11:13

hmakholm left over Monica

291,611

What is vacuously false here when $B$ is $\emptyset$? – Error 404 Jul 27 '17 at 11:34
1

@Mathmore: "There is an $x\in\varnothing$ such that ..." is vacuously false. – hmakholm left over Monica Jul 27 '17 at 11:44
Gotcha! ${}{}{}{}$ – Error 404 Jul 27 '17 at 11:58

Sahiba Arora · Accepted Answer · 2017-07-27T11:24:16.037

2

Second statement shows that $$\emptyset \subseteq A^c$$ which is also true. We are not preferring one over other. Both are true.

Edit 1: The second statement does not imply that $\emptyset$ is not a subset of $A$. For that you would need that there exists $x \in \emptyset$ such that $x \in A$. This is indeed false.

Edit 2: What second statement implies is $\emptyset \cap A=\emptyset.$

edited Jul 27 '17 at 11:24

answered Jul 27 '17 at 11:08

Sahiba Arora

11,065

So the statement "empty set is not a subset of every set" is also true? – Error 404 Jul 27 '17 at 11:11
That is not what the second statement implies. – Sahiba Arora Jul 27 '17 at 11:12
If empty set is a subset of $A^c$ then it is surely not a subset of $A$. Is this argument wrong? – Error 404 Jul 27 '17 at 11:18
3

@Mathmore yes this wrong. But it is only wrong for the empty set. – quid Jul 27 '17 at 11:18
@Mathmore Check my edit. – Sahiba Arora Jul 27 '17 at 11:24
1

@Mathmore, no. $X \subseteq Y, X \subseteq Y^c \implies X \subseteq Y \cap Y^c=\emptyset.$ Hence, $X=\emptyset$. – Sahiba Arora Jul 27 '17 at 11:40
@quid I learnt definition of $A^c$ as a set of those elements which do not belong to $A$. – Error 404 Jul 27 '17 at 11:40
Yes, that is the definition. – Sahiba Arora Jul 27 '17 at 11:41
1

@SahibaArora Awesome! +1 – Error 404 Jul 27 '17 at 11:43

score -1 · Answer 3 · answered Jul 27 '17 at 11:19

In elementary set theory, each set can be defined as $A=\{x\mid P(x)\}$ where $P(x)$ is a predicate with free variable $x$, and each subset as $B=\{x\mid x\in A\wedge Q(x)\}$ with predicate $Q$.

Define $\emptyset_A=\{x\mid x\in A\wedge x\not\in A\}$. This is a subset of $A$, the empty subset of $A$. For any other set $B$, define similarly $\emptyset_B=\{x\mid x\in B\wedge x\not\in B\}$ as the empty subset of $B$.

But $\emptyset_A=\emptyset_B$, since $x\in\emptyset_A\Leftrightarrow x\in A\wedge x\not\in A\Leftrightarrow x\in B\wedge x\not\in B\Leftrightarrow x\in\emptyset_B$. In the 2nd equivalence, a false assertion is replaced by another false assertion.

Hence, the empty subset of a set $A$ is independent of the set $A$ itself and so we can safely speak of the empty set by putting $\emptyset=\emptyset_A$.

What is this "elementary set theory"? Also, $P(x)$ given by $x\notin x$ or $x=x$ will probably cause you some issues with that. — Asaf Karagila, Jul 27 '17 at 11:20

Confusion in understanding why empty set is a subset of every set

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