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 November 11th, 2017, 06:28 AM #1 Member   Joined: Oct 2017 From: Rumba Posts: 34 Thanks: 0 Set notation with empty set Let A = ∅, B = {7, 8, 9, 1, {∅}}. For each of the following, state whether the statement is True or False. (a) ∅ ∈ A = true (b) A ⊂ B = false (c) P(A) ∈ B = false (d) B ⊆ A = true (e) A ⊆ B = true (f) A ⊆ ∅ = false
November 11th, 2017, 09:48 AM   #2
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Quote:
 Originally Posted by sita Let A = ∅, B = {7, 8, 9, 1, {∅}}. For each of the following, state whether the statement is True or False. (a) ∅ ∈ A = true
No, the empty set is empty. It does not have any elements. This is false.

Quote:
 Originally Posted by sita (b) A ⊂ B = false
No, the empty set is a subset of every set, so this is true.

Quote:
 Originally Posted by sita (c) P(A) ∈ B = false
No. What is $P(A)$? The power set of the empty set is $\{\emptyset\}$, which is an element of B. This is true.

Quote:
 Originally Posted by sita (d) B ⊆ A = true
No, this is false. It is not the case that every element of B is an element of A.

Quote:
 Originally Posted by sita (e) A ⊆ B = true
Yes. The empty set is a subset of every set.

Quote:
 Originally Posted by sita (f) A ⊆ ∅ = false
Yes. Every set is a subset of itself.

Last edited by Maschke; November 11th, 2017 at 09:51 AM.

 November 11th, 2017, 10:00 AM #3 Math Team     Joined: May 2013 From: The Astral plane Posts: 1,914 Thanks: 774 Math Focus: Wibbly wobbly timey-wimey stuff. I don't want to distract from the OP's problem, but I'm curious.... Don't a) and f) contradict each other? If $\displaystyle A \in \emptyset$ because every set is a subset of itself, then why isn't $\displaystyle \emptyset \in A$ by the same argument? -Dan
November 11th, 2017, 10:47 AM   #4
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Quote:
 Originally Posted by topsquark I don't want to distract from the OP's problem, but I'm curious.... Don't a) and f) contradict each other? If $\displaystyle A \in \emptyset$ because every set is a subset of itself, then why isn't $\displaystyle \emptyset \in A$ by the same argument? -Dan
The empty set is empty, so $x \in \emptyset$ is false for every $x$.

But the empty set is a subset of every set, including itself. We say $X \subset Y$ if it is the case that $x \in X \implies x \in Y$. For any set $X$, it's vacuously true that if $x \in \emptyset$ then $x \in X$, because there is no $x$ that's an element of $X$.

In other words the statement $x \in \emptyset \implies x \in X$ is true, because the antecedent is always false. $x \in \emptyset$ is always false, making the implication true. So by definition, $\emptyset \subset X$.

So the empty set contains no elements, but the empty set is a subset of every set.

Last edited by Maschke; November 11th, 2017 at 10:57 AM.

 November 11th, 2017, 06:13 PM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 The difference between (a) and (f) is the difference between "$\displaystyle \in$" and "$\displaystyle \subset$" Thanks from Maschke and topsquark

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