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well I guess it would be easier to show an example of what I am asking:

Is this sentence True or False ?

$$\{\emptyset\}\subseteq \{1,\{\emptyset\}\}$$

Edit:

following the same logic I had before I got the answer to the original question in the post I tried to solve this 2 questions : edit - 2 questions photo - please edit to show it to everyone

and on the left question I figured the answer was True

and on the right question I figured the answer was True also

but after reading the answers it seems to me that it cannot be that both are true but I can't explain why... Am I correct on this one ?

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  • $\begingroup$ Please make the image inline by putting "!" before the square brackets of example. $\endgroup$
    – Kumar
    Nov 18, 2019 at 18:40
  • $\begingroup$ @Kumar New users do not have the ability to place inline images. This is in an attempt to curb the effect of spam bots and trolls among other things. $\endgroup$
    – JMoravitz
    Nov 18, 2019 at 18:41
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    $\begingroup$ The end result for the mathematics involved here is to recognize that the "depth" here matters, the number of braces... Each of $\emptyset,\{\emptyset\},\{\{\emptyset\}\},\dots$ are distinct, different, nonequal sets. Then, recognize that $a\in A$ is asking specifically if $a$ is an element of $A$ which is different than asking if $a$ is an element of an element of $A$ or deeper, i.e. that it is asking if the element $a$ appears "at the topmost level." Similarly, $A\subseteq B$ is asking if every element of $A$ (at the topmost level) is also an element (at the topmost level) of $B$ $\endgroup$
    – JMoravitz
    Nov 18, 2019 at 18:46
  • $\begingroup$ In your case, $\emptyset$ is an element of the set on the left but is not an element of the set on the right in exactly the same way that $\{3\}$ is not a subset of $\{1,\{3\}\}$. $\endgroup$
    – JMoravitz
    Nov 18, 2019 at 18:47
  • $\begingroup$ I think I got it , although {} is still a litlle bit tricky for me. thanks guys $\endgroup$
    – Maor
    Nov 18, 2019 at 18:57

2 Answers 2

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You ask if this is correct:

$$\{\emptyset\}\subseteq \{1, \{\emptyset\}\}$$

It is not. Indeed, $\emptyset \in \{\emptyset\}$ while $\emptyset \notin\{1, \{\emptyset\}\}$.

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  • $\begingroup$ thanks for the fast reply ! $\endgroup$
    – Maor
    Nov 18, 2019 at 18:55
  • $\begingroup$ You are welcome! $\endgroup$
    – J. De Ro
    Nov 18, 2019 at 19:11
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There are only 2 cases :

(1) either all the elements of the set on the left are also elements of the set on the right, and in that case , the statement is true ( set on the left is included in set on the right)

OR

(2) at leat one element of the set on the left is not an element of the set on the right.

Here we are on case 2. Indeed I can exhibit the element " empty set" that belongs to the set on the left but not to the set on the right.

Note: the empty set is a set, but it can also be an element of a set; that is what happens here; and, in fact, it is the only element of the set on the left.

The set on the left can be compared to a box with an empty box inside. The empty box is empty. But the box with an empty box inside is not empty, it has 1 element ( namely, the empty box).

enter image description here

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  • $\begingroup$ but if you mark {empty set} = A then isn't the right side says : {1,A} ? $\endgroup$
    – Maor
    Nov 18, 2019 at 21:59
  • $\begingroup$ @Maor.- I've added a picture to my post. Hope it helps. $\endgroup$
    – user654868
    Nov 18, 2019 at 22:45

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