True or false - Set theory

[Randyyy]

set A = {0,{1},{},{2,3}} which of the following statements are true or false?


1: True because Null is a subset of all sets.
2: True, because Null is an element contained in set A.
3: false because 1 is not a set to begin with so it is unable to be a proper subset.
4: True, because the set containing the element 1 is an element of set A.
5: False because the element 1 is contained within a set of A and not within A itself.
6: False because the set {{1}} does not contain all elements of A.
7:False because 1,2,3 are not elements of set A, rather they are contained in sets within the set of A.
8: False because only 0 is an element of set A.

is my reasoning correct or am I overlooking something?
Thanks in advance!

 

[pka]

set A = {0,{1},{},{2,3}} which of the following statements are true or false?
View attachment 21424
1: True because Null is a subset of all sets.
2: True, because Null is an element contained in set A.
3: false because 1 is not a set to begin with so it is unable to be a proper subset.
4: True, because the set containing the element 1 is an element of set A.
5: False because the element 1 is contained within a set of A and not within A itself.
6: False because the set {{1}} does not contain all elements of A.
7:False because 1,2,3 are not elements of set A, rather they are contained in sets within the set of A.
8: False because only 0 is an element of set A.
is my reasoning correct or am I overlooking something?

$\{1\}\in A~\therefore~\{\{1\}\}\subseteq A$

 

[Randyyy]

$\{1\}\in A~\therefore~\{\{1\}\}\subseteq A$

Isn´t it so that {{1}} ⊂ A because all elements of the set is contained in A but A has got more elements than that set?

 

[pka]

Isn´t it so that {{1}} ⊂ A because all elements of the set is contained in A but A has got more elements than that set?

No you have that wrong. The symbol $\subseteq$ is read "is a subset of"
The symbol $\subset$ is read "is a proper subset of"
So $\{1,2\}\subset \{1,2,3\}$ as well as $\Large\{1,2\}\subseteq \{1,2,3\}$

 

[HallsofIvy]

(6) does not assert that all elements of A are in {{1}}, just that all elements of {{1}}, and that is only {1}, are in A

 

[Randyyy]

No you have that wrong. The symbol $\subseteq$ is read "is a subset of"
The symbol $\subset$ is read "is a proper subset of"
So $\{1,2\}\subset \{1,2,3\}$ as well as $\Large\{1,2\}\subseteq \{1,2,3\}$

ohh, so I got my definitions flipped. So then (6) is a subset of A. But isn´t (1) false then because the definition says that null is a subset of all sets and not a proper subset?

(6) does not assert that all elements of A are in {{1}}, just that all elements of {{1}}, and that is only {1}, are in A

ahh yeah, I see what you are saying. I had proper subset and subset confused, for some reason I keep confusing myself whenever I am to decide which is which.

 

[pka]

But isn´t (1) false then because the definition says that null is a subset of all sets and not a proper subset?

Whether or not $\emptyset$ is a proper subset of a set depends upon the textbook in use.
You must follow your textbook.

 

[Randyyy]

Whether or not $\emptyset$ is a proper subset of a set depends upon the textbook in use.
You must follow your textbook.

I checked and the definition in my textbook says that ∅ is a subset of all sets. Going with that definition the answer should then all be the following:
1: False because Null is a subset of all sets and not a proper subset of all sets..
2: True, because Null is an element contained in set A.
3: false because 1 is not a set to begin with so it is unable to be a subset of A.
4: True, because the set containing the element 1 is an element of set A.
5: False because the element 1 is contained within a set of A and not within A itself.
6: True because all elements of the set containing {{1}} is an element of the set A.
7:False because 1,2,3 are not elements of set A, rather they are contained in sets within the set of A.
8: False because only 0 is an element of set A.

 

[pka]

Well if a student in one of my classes marked #1 false I would mark his/.her answer wrong.
Most working mathematicians accept Paul Halmos' book Naive Set theory as the ultimate guide.
He says that the emptyset is a proper subset of every set. On page 3, he writes that if $A\subseteq B$ and $A\ne B$ then $A$ is a proper subset of $B$.
SEE HERE

 

[Randyyy]

Well if a student in one of my classes marked #1 false I would mark his/.her answer wrong.
Most working mathematicians accept Paul Halmos' book Naive Set theory as the ultimate guide.
He says that the emptyset is a proper subset of every set. On page 3, he writes that if $A\subseteq B$ and $A\ne B$ then $A$ is a proper subset of $B$.
SEE HERE

I checked out the wiki page you attached and found the following definition: "A special case is the empty set { }, denoted by ∅. It is also a subset of any given set X. It is also always a proper subset of any set except itself."

So it does seem like whether or not #1 asked if ∅ was a subset or proper subset of A if should always be true per definition.

 

[Steven G]

@pka What might be wrong with the following proof (I think it is fine)
Fact: Given any set A, $$∅ \subseteq A$$Proof: Suppose $$∅ \nsubseteq A$$. Then exists x in ∅ with x not in A. But this is absurd as there is no element in the empty sen. Therefore $$∅ \subseteq A$$

 

[pka]

@pka What might be wrong with the following proof (I think it is fine)
Fact: Given any set A, $$∅ \subseteq A$$Proof: Suppose $$∅ \nsubseteq A$$. Then exists x in ∅ with x not in A. But this is absurd as there is no element in the empty sen. Therefore $$∅ \subseteq A$$

I don't really follow your point. The statement that $A$ is a proper subset of $B$ means $A\subseteq B\wedge A\ne B$.
Is the empty set a proper subset of every set?

 

[Steven G]

I don't really follow your point. The statement that $A$ is a proper subset of $B$ means $A\subseteq B\wedge A\ne B$.
Is the empty set a proper subset of every set?

Come on that is the typical proof to show that the empty set is a subset of every set.

You said that the empty set may not be a subset of every set depending on which book you use. I am asking how that can be if I proved the statement. The only way an author could say otherwise is if my proof is wrong. So where did I go wrong? BTW, I know you follow Halmos and think that in fact the empty set is a subset of every set.