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According to the definition of empty set that is the set containing no elements, commonly denoted emptyset or emptyset , Really i'm confused the empty set is nothing and hasn't any element and it is used widely in mathematics especially topology and algebra group , Now my question here is :

Question: let $A$ be a set of elements and ${\emptyset}‎$ is empty set , Why : ‎ ‎$\{A,\emptyset\}\ and \{A\}$ haven't the same meaning ?

Not: if they have the same meaning then why should be cite the empty set as a condition in the definition of topological space .

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    $\begingroup$ The fist of those two sets has two elements, the second one has one element. They "don't have the same meaning". $\endgroup$
    – Git Gud
    Commented Feb 11, 2017 at 19:06
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    $\begingroup$ Typically I'm against using real world examples to explain mathematics, specially something like set theory. But here it is anyway. Suppose you have two boxes, box $A$ has some stuff, box $\varnothing$ is empty. Is holding both boxes the same as holding just box $A$? $\endgroup$
    – Git Gud
    Commented Feb 11, 2017 at 19:10
  • $\begingroup$ What you have si a set: $\{A,\emptyset\}$, the set that contains two sets, $A$ and $\emptyset$ not $\{A \cup \emptyset\}$ $\endgroup$
    – MasB
    Commented Feb 11, 2017 at 19:10
  • $\begingroup$ @GitGud, your example convinced me , i taught Empty set that is nothing as the meaning of 0 as a number $\endgroup$
    – zeraoulia rafik
    Commented Feb 11, 2017 at 19:19
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    $\begingroup$ "i'm confused the empty set is nothing" The empty set is not nothing. The empty set is a set that has nothing in it. A notebook with nothing written in it is not nothing, it is a notebook. A drawer with a can opener in it is not a can opener. It has a can opener in it but it is not a can opener itself. A set is not the same thing as the things inside it. An empty set is a thing-- a thing that has nothing in it. So {A, $\emptyset$} is a set with two things in it. One thing is A. The other is the emptyset. {A} is a set with one thing in it. The thing is A. The empty set is not in it. $\endgroup$
    – fleablood
    Commented Feb 11, 2017 at 19:19

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No, they have not: the set $\{A,\varnothing\}$ has two elements unless $A=\varnothing$, whereas the sets $\{A\}$ and $\{\varnothing\}$ have one.

In particular, if $A=\{\varnothing\}$, the set $\;\{\{\varnothing\},\varnothing\}$ has two elements – this is even von Neumann's definition of the number $2$.

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$\{ A , \emptyset \}$ does not mean $A \cup \emptyset$.

$\{ A , \emptyset \}$ is the set containing the two elements $a,b$, namely $a = A$ and $b = \emptyset$.

$\{A\}$ does not mean $A$, it means the set containing $A$ as an element.

It could also happen that $A =\emptyset$, in which case $\{A, \emptyset\} = \{A\} = \{\emptyset\}$, and the two sets have the same meaning.

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If $A$ is not the empty set, then the meaning is different because $\varnothing\in\{A,\varnothing\}$ is true and $\varnothing \in \{A\}$ is false. Sets are by definition different when there's something that is an element of one but not of the other.

If $A$ is the empty set, then $\{A,\varnothing\}$ and $\{A\}$ are the same; they are then just different ways to write the set also known as $\{\varnothing\}$.

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  • $\begingroup$ well , pleas read the not that i wrote below of the question , then if they are both the same why should cite empty set in the definition of topological space since it's trivial satisfied ? $\endgroup$
    – zeraoulia rafik
    Commented Feb 11, 2017 at 19:09
  • $\begingroup$ @zeraouliarafik: I do not understand what you're trying to ask there. $\endgroup$
    – hmakholm left over Monica
    Commented Feb 11, 2017 at 19:10
  • $\begingroup$ In my mind every non -empty set contain empty set $\endgroup$
    – zeraoulia rafik
    Commented Feb 11, 2017 at 19:11
  • $\begingroup$ @zeraouliarafik: The answer says they are not the same. $\endgroup$
    – Eric Wofsey
    Commented Feb 11, 2017 at 19:12
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    $\begingroup$ @zeraouliarafik: Every set has the empty set as a subset. Most sets that you meet in practice do not have the empty set as an element. Being an element is something completely different from being a subset. It is regrettable and confusing that the word "contains" is being used for both of these; it is best to avoid the word "contains" completely until you are completely sure how sets work. $\endgroup$
    – hmakholm left over Monica
    Commented Feb 11, 2017 at 19:14
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Babar, Haitha, Tantor and Pink Honk-Honk are four elephants.

A = $\{$Babar, Haitha, Tantor, Pink Honk-honk$\}$.

A is one thing. It is a set that contains four elephants. But A is not an elephant. A is not even four elephants really. It is a collection of four elephants.

A has four things in it. But A, itself, is one thing.

B = {A}={{Babar, Haitha, Tantor, Pink Honk-honk}} has one thing in it. It has a set in it. It does not have four elephants in it. It is a set with only one thing in it. That thing is a set. (B has something inside it that has four elephants but B, itself, does not have any elephants directly inside it.)

{A} $\ne $ A.

$\emptyset = \{\}$ is set with nothing in it. It is something. It is a set. It is a set with nothing in it.

{A} is a set with one thing in it. {A, $\emptyset$} is a set with with two things in it.

{A} $\ne $ {A, $\emptyset$} because the two sets have different things inside them.

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Note: $A \cup \emptyset = A$ but $\{A, B\} \ne \{A \cup B\}$ so $\{A\} = \{A \cup \emptyset\} \ne \{A, \emptyset\}$.

$\{A,B\}$ is a set with two things in it. Those two things are two sets. $A \cup B$ is one set; a combined set-- all the elements of A and B are dumped out, mixed together and packed up into a new set. So $\{A\cup B\}$ is a set with only one thing int it. That one thing is one set. So $\{A,B\} \ne \{A \cup B\}$.

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