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I came across this question in my assignment.

Question 1. Let $A=\{\emptyset,\{\emptyset\},1,\{1,\emptyset\},2\}$. Does the following hold true for the above set?

$B=\{\emptyset,\{\emptyset\},\{1,\emptyset \} \} \subset A$ ? (It is told to be true in the soultions but I think it is not.)

I thought about it. Compared it with the set A but that took me nowhere. What I think is that it is false because...

1. In order to be a subset of a set (A here), the "original elements" are taken out from the "original set" (again A) and are put into these $\{$ $\}$ which converts a respective element into a subset of the set.

Now if I compare set A and the other set B (which I have to tell about if it holds true or not with respect to set A), I find...

(A). $\emptyset$ is a subset of every set so B should have $\emptyset$ contained in it. (Yes!)

(B). $\emptyset$ is taken out from set A and has been placed in set B so it should be in brackets. (yes it is!) like this $\{\emptyset\}$

(C). Coming to third and the last element of set B which is $\{1,\emptyset\}$, I find 1 when taken out from set A should simply have been placed like this $\{1\}$ but it is not rather it has been placed like this $\{1,\emptyset\}$ which I think is incorrect. (But soluions say it is correct, which I don't agree to).

(D). If they were have to be placed in B then I think there could have been a better way which is this... $B=\{\emptyset,\{\emptyset\},\{1\},\{\{\emptyset \}\} \} \subset A$ (True?) Or this $A=\{\emptyset,\{\emptyset\},1,\{\{1,\emptyset\}\},2\}$ (True?)

Am I correct in my approach? Also If you would like to suggest me something, go ahead.

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    $\begingroup$ For each element of $B$, see if it's an element of $A$. It's that simple. $\endgroup$ – quasi Nov 6 '17 at 8:37
  • $\begingroup$ What is the answer finally? I want to know this. Actually. $\endgroup$ – Saksham Sharma Nov 6 '17 at 8:38
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    $\begingroup$ $B$ is a proper subset of $A$, so $B \subset A$. $\endgroup$ – quasi Nov 6 '17 at 8:38
  • $\begingroup$ You did not understand. Just answer this. Is it $true$ or $false$? $\endgroup$ – Saksham Sharma Nov 6 '17 at 8:40
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    $\begingroup$ @Raffaele The case $B=\emptyset$ would be actually redudnant in that definition (and the $\Rightarrow$ would be syntactically wrong). $\endgroup$ – user228113 Nov 6 '17 at 9:55
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The set $B$ has $3$ elements, namely $$\emptyset,\;\;\{\emptyset\},\;\;\{1,\emptyset \}$$ The set $A$ has $5$ elements, namely $$\emptyset,\;\;\{\emptyset\},\;\;1,\;\;\{1,\emptyset\},\;\;2$$ Since each element of $B$ is also an element of $A$, we have $B \subseteq A$.

But $B \ne A$, hence $B \subset A$.

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  • $\begingroup$ By the way, for many authors $\subset$ and $\subseteq$ are interchangeable. $\endgroup$ – littleO Nov 6 '17 at 8:47
  • $\begingroup$ Right, but most modern texts (e.g., last $20$ years) appear to have chosen to $\subset$ to mean proper subset, so the notation is evolving. $\endgroup$ – quasi Nov 6 '17 at 8:48
  • $\begingroup$ The only potential trouble here is that if we're working in one of the (common!) developments of axiomatic set theory where $1=\{\varnothing\} $ and $2=\{1, \varnothing\} $, in which case $B=A$ and it then depends on notational convention whether we can say that $B\subset A$. $\endgroup$ – Henning Makholm Nov 6 '17 at 8:56
  • $\begingroup$ Agreed, but then that context would need to be specified. I'm assuming a more elementary context (e.g,, a first introduction to sets). $\endgroup$ – quasi Nov 6 '17 at 9:01
  • $\begingroup$ @quasi If we want to take out some elements from a set A$=\{1,2,\{3,4\},5\}$ and write them as subset of set A then won't each element come in brackets and then written as set B$=\{\{1\},\{2\}\}$. Here is A $\subset$ B? Or to be a subset simply the subset should contain elements of the universal set and put them under initial and last bracket? $\endgroup$ – Saksham Sharma Nov 6 '17 at 10:55
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To check if $B$ is a subset of $A$, we can just check each element of $B$ individually, and see if each element of $B$ is also an element of $A$.

By the way, how many elements does $B$ have? It has three elements. And what are they? Well, they are: $\emptyset, \{\emptyset\}$, and $\{1,\emptyset\}$.

Let's check those three elements one by one. We can start with $\{1,\emptyset\}$. Is this an element of $A$? The answer is yes. The set $A$ has five elements, and one of them is $\{ 1, \emptyset \}$.

You can check the other elements of $B$ similarly.

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  • $\begingroup$ If we want to take out some elements from a set A$=\{1,2,\{3,4\},5\}$ and write them as subset of set A then won't each element come in brackets and then written as set B$=\{\{1\},\{2\}\}$. Here is A $\subset$ B? Or to be a subset simply the subset should contain elements of the universal set and put them under initial and last bracket? $\endgroup$ – Saksham Sharma Nov 6 '17 at 10:58

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