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 November 11th, 2017, 05:28 AM #1 Member   Joined: Oct 2017 From: Rumba Posts: 39 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, 08:48 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 08:51 AM. November 11th, 2017, 09:00 AM #3 Math Team   Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 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, 09:47 AM   #4
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 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 09:57 AM. November 11th, 2017, 05:13 PM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The difference between (a) and (f) is the difference between "$\displaystyle \in$" and "$\displaystyle \subset$" Thanks from Maschke and topsquark Tags check, empty, notation, set Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Lalitha183 Abstract Algebra 1 June 5th, 2017 07:00 AM GIjoefan1976 Algebra 49 April 7th, 2017 12:34 PM idontknow Physics 16 March 8th, 2016 10:29 PM barokas Applied Math 4 September 25th, 2013 03:47 PM outsos Applied Math 36 April 30th, 2010 10:46 AM

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