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This problem is from Discrete Mathematics and its Applications enter image description here

My question is on 9b. I know that the sign represents an element is a member of. (from book) enter image description here

I know that the O with a slash across it is the empty set which "is a special set that has no elements". From http://mathcentral.uregina.ca/QQ/database/QQ.09.06/narayana1.html, I got that the empty set is a subset of all sets, meaning that every member of the empty set(nothing) is also a member of any other set.

Based on all of this, for 9b, would {0} contain the empty set because it fundamentally has the elements that consist of the empty set(nothing) or does it physically have to have the empty set?

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    $\begingroup$ $\varnothing \subset A$ for every $A$ (including $\varnothing$) but it is not true that $\varnothing\in A$ for every $A$. $\endgroup$ – Hugh Jan 14 '15 at 7:49
  • $\begingroup$ because A contains at least every element in the empty set(nothing) $\endgroup$ – committedandroider Jan 14 '15 at 7:52
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    $\begingroup$ Or, put another way, there is no element of $\varnothing$ that isn't also an element of $A$; (it's also the case there is no element that isn't also not an element of $A$, but that's not relevant for determining set inclusion). $\endgroup$ – Hugh Jan 14 '15 at 7:54
  • $\begingroup$ i mean that doesn't really tell you anything though. If p is an element in ∅ is not in A and q is ∅ is subset of A, then you can form implication p -> ~q, you don't know anything if p is false, that is there is no such element in ∅ that is not in A. $\endgroup$ – committedandroider Jan 14 '15 at 8:10
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$x \in \{ y \} $ if and only if $x = y$. Thus, $\varnothing \in \{ 0 \}$ if and only if $\varnothing = 0$.

Out of context, there is actually ambiguity here. Often, in set-theoretic contexts, we interpret natural numbers as being the set of all smaller natural numbers; e.g. $3 = \{ 0, 1, 2 \}$. And according to this convention, $0$ is indeed equal to $\varnothing$.

But we might not adopt this convention, and we take $0$ to be its own thing that is unequal to $\varnothing$ or any other set that is 'naturally' written.

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    $\begingroup$ for 3 = {0,1,2}, do you mean the size of that set is 3? $\endgroup$ – committedandroider Jan 14 '15 at 6:51
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    $\begingroup$ 3 is the set of {0, 1, 2}? I don't quite get taht $\endgroup$ – committedandroider Jan 27 '15 at 6:09
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When $X$ and $Y$ are two sets, we say that $X\subset Y$ if every element of $X$ is contained in $Y$.

With this definition, you see that $\emptyset \subset Y$ for any set $Y$. Indeed, there is no element in $\emptyset$, so every element of $\emptyset$ is contained in $Y$ (trivially true as there is nothing to check).

However, if you want to write $\emptyset \in Y$, this means that there is one element of $Y$ which is a set and that this set is the empty set. When $Y=\{0\}$, you have only one element in $Y$, and this one is not a set, it is a number, which is $0$. Hence, $\emptyset\notin \{0\}$.

Both statements $9a$ and $9b$ are false.

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  • $\begingroup$ Yeah its weird because if Y contained say {1,2,3,4}. It doesn't make sense to me to say it also contains nothing(content of empty set) because its a contradiction to what it actually contains(some stuff) $\endgroup$ – committedandroider Jan 14 '15 at 6:53
  • $\begingroup$ When we say that $Y$ contains something, you should think as "$Y$ contains at least this something" (sorry for my approximative English). Hence, the sentence "$Y$ contains at least nothing" becomes more intuitive. $\endgroup$ – Jérémy Blanc Jan 14 '15 at 6:56
  • $\begingroup$ Yeah thank you. that makes so much more sense. It contains at least nothing because it contains all those elements :) No worries. $\endgroup$ – committedandroider Jan 14 '15 at 6:57
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    $\begingroup$ I think this should be the best answer ! $\endgroup$ – virat Mar 25 '17 at 7:44
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There are several ways to represent the empty set. $\{ \}, \emptyset, \text{ and } \varnothing$ are three common ways.

Saying "the empty set(nothing)" is incorrect. The empty set is the set that contains nothing. A bottle can contain nothing, but the bottle itself is something.

Hence, for example, the set $\{\varnothing\}$ is not the empty set simply because it has something in it. In English, the set containing the empty set is not the empty set.

For the empty set to be a member of a set, it has to actually be in that set. The empty set is in $\{1,2,\varnothing\}$. The sets $\{1,2\}$ and $\{1,2,\{\varnothing\}\}$ do not have the empty set in them.

A subset of the set S, is either the set S or the set S with some stuff removed from it.

For example, a subset of $\{a,b\}$ is the set $\{a,b\}$ with $0$ to $2$ things removed from it. These sets are subsets of $\{a,b\}$:

\begin{align} \{a,b\} &- \text{nothing was removed}\\ \{b\} &- \text{a was removed}\\ \{a\} &- \text{b was removed}\\ \{ \} &- \text{a and b were removed} \end{align}

where the last set, $\{ \}$, is the empty set.

Start with any set, take everything out if it, and you are left with an empty set. Hence the empty set is a subset of every set.

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No. Elements and subsets are not the same thing.

An element of a set is one of the things in the set.

A subset of a set is another set that does not contain any elements which are not elements of the set to which it is a subset.

The empty set is not an element of {1,2,3}.

That is because 1 is not the empty set, 2 is not the empty set, and 3 is not the empty set.

"Empty set" does not mean "nothing". Instead, it is the set that does not have elements.

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Recall the (naive) definitions of the symbols $\subset$ and $\in$. The reason that $\varnothing\subset A$ for any set $A$ is because any "$x$ in $\varnothing$" is automatically also in $A$ (vacuously, because there is no such $x$).

On the other hand, $\varnothing$ is an element of, say, the set $\{ \varnothing\}$, for the same reason that $1\in \{1\}$.

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  • $\begingroup$ I mean if you had a set of cars, say 2014 Toyota Rav 4, 2013 Toyota Rav 4, 2012 Rav 4, it just doesn't make sense to me to say that this set also contains nothing(content of the empty set) because it is a contradiction of what it actually is (contains some stuff) $\endgroup$ – committedandroider Jan 14 '15 at 6:55
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Lets consider this situation: Let set $A = {1,2,3}$, if null/empty set is an element of any set, therefore the cardinality of set A is 4 which is not true from the definition of cardinality of set which states that it is the number of elements in a given set. The truth is, set A has a cardinality of 3 since it has only three elements and it has 8 subsets. In other words, null set is not an element of any set but a subset of any set.

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  • $\begingroup$ This is an old question which already has a well-accepted answer. You have contributed nothing new. Please refrain from answering such old questions $\endgroup$ – Shailesh Feb 23 '16 at 1:52

protected by Alex M. Sep 7 '16 at 16:48

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