0
$\begingroup$
  1. True or false. A set is any collection of objects.

  2. True or false. A proper subset of a set is itself a subset of the set, but not vice versa.

  3. True or false. The empty set is a subset of every set.

I need you confirm if my answers are correct or wrong

my effort

1-TRUE. A set is a collection of objects. The objects are referred to as elements of the set.

2-FALSE. A “proper subset” of a set A is defined as a set B that is contained by A, but is not equal to A. If A had a subset B, where B is defined as A, then A=B, and thus does not satisfy the conditions for a proper subset, although it is still always a subset of itself.

3-TRUE. An empty set contains no elements while a subset contains elements that are not in the other comparing set. Hence an empty set becomes a subset of all the other sets because it has no elements and the other set contains elements.

$\endgroup$
  • $\begingroup$ If you want to be more precise, check out Russell's paradox. $\endgroup$ – Michael Burr Aug 15 '15 at 12:41
  • $\begingroup$ Have you considered the collection of all sets? (in regard to the 1st part). $\endgroup$ – Gerry Myerson Aug 15 '15 at 12:42
  • $\begingroup$ You write "2-FALSE", but follow that with a correct argument that claim 2 is true. Is the "false" a typo? $\endgroup$ – hmakholm left over Monica Aug 15 '15 at 13:02
0
$\begingroup$
  1. FALSE

A set can't be any collection of objects because certain collections are paradoxical. For example, take that a set is a collection of all sets such that these sets are not members of themselves (Russel's Paradox), and we'll call that set $A$. Then by definition $A$ should include itself in $A$ but if it does so then $A$ contains a member(itself) that is a member of itself. And if $A$ does not include itself then it can't be a set of all set such that these sets do not contain themselves.

Another simpilier paradox is Cantor's paradox in naive set theory that says say that $X$ is the set of all sets, but such a set can't exist because you can always formulate a new set that contains all the elements that $X$ does and the set including $X$ that is ${X}$, so you can't have a set of all sets.

2.TRUE If $A$ is equal to $B$ then all the elements in A are also in B and vice verse. But if $A$ does not contain all the elements $B$ or $B$ does not contain all elements in $A$ only then can one a be a proper subset of the other. So it is true.

  1. TRUE The empty set is a subset of every set. This has to do with the definition of a subset.

Please see the link below if you want to see an explanation

http://mathcentral.uregina.ca/QQ/database/QQ.09.06/narayana1.html

and here if want to see the proof

https://books.google.com/books?id=NuFeW8N2hlkC&pg=PA21&lpg=PA21&dq=proof+the+O+is+subset+of+every+set&source=bl&ots=5TpYc_2MEb&sig=WNghOHNFCJHdw-mlynVH9Dle-uY&hl=en&sa=X&ved=0CFIQ6AEwCGoVChMIsumpqcmrxwIVxRw-Ch2fYw1C#v=onepage&q=proof%20the%20O%20is%20subset%20of%20every%20set&f=false

$\endgroup$
0
$\begingroup$

FALSE. If you allow any set, you end up with paradoxes. The most famous of these is Russell's paradox.

TRUE.

TRUE. For every element in the empty set, that element is an element of any set $A$. This is vacuously true.

$\endgroup$
  • $\begingroup$ what is the reason for 2 to be true ?? I need reason not just true of false $\endgroup$ – Adel Hassan Aug 15 '15 at 12:46
  • $\begingroup$ If $A \subset B$ then every element of $A$ is contained in $B$. However, not every element of $B$ needs to be contained in $A$ since you said proper subset. For example, take your sets to be $\Bbb{R}$, the real numbers, and $\Bbb{N}$, the natural numbers. Then it is true that $\Bbb{N} \subset \Bbb{R}$ but not vice versa. $\endgroup$ – Race Bannon Aug 15 '15 at 12:57
0
$\begingroup$

You cannot just take any collection of objects to be a set.

There has been a long process in clarifying set theory until it is consistent.

There are several set theories in fact in modern mathematics.

The most commonly used is the set theory given by adopting the Zermelo-Fraenkel axioms.

Here they are http://mathworld.wolfram.com/Zermelo-FraenkelAxioms.html

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.