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My Discrete Mathematics book talks about the proof of why null set is the subset of every non-empty set. It has raised some questions in my head.

Firstly, Why? I mean why will we want to make simple, intuitive things so complicated. I mean why is there need to say tht non-empty set contains an empty set? We can see that a non-empty set has elements and combination of atleast one of them will be a subset of the main set which makes perfect sense. But why will we need to bring up the whole counter intutive idea of null set being a subset? Same question about the set being the subset of itself.

What I am looking for is that if somebody can help me understand the purpose and importance of doing so.

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    $\begingroup$ I'm confused: Are you saying that $\emptyset\subseteq A$ is so simple and intuitive that it need no complicated proof, or are you saying that $\emptyset\subseteq A$ is counterintuitive? $\endgroup$ – Hagen von Eitzen Sep 20 '15 at 21:53
  • $\begingroup$ What is so counter intuitive? By adding elements to the empty set, you can create any set. By removing elements from any set, eventually you will get to the empty set. Wouldn't it be strange to define things otherwise? $\endgroup$ – DanielV Sep 20 '15 at 22:09
  • $\begingroup$ A subset of some main set is a combination of at least zero elements of that set. $\endgroup$ – user253751 Sep 21 '15 at 2:39
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What you write is not true at all. Take $X=\{\varnothing\}$, and look at $Y=\{X\}$. The set $Y$ has one element, which is $X$. So $Y$ has one element which is neither empty, nor a subset of $Y$.

On the other hand, $\varnothing\subseteq Y$, because every element of $\varnothing$ is an element of $Y$. Simply because $\varnothing$ has no elements!

There are three main reasons to prove that $\varnothing\subseteq A$ for any set $A$:

  1. It is an excellent opportunity to discuss "vacuous truth" or "vacuous implications", where something is true simply because it cannot be false (i.e., there are no counterexamples).

  2. It establishes that $\varnothing$ is the "smallest" possible set, and that it is unique (if $A$ and $B$ are two empty sets, then $A\subseteq B$ and $B\subseteq A$, so $A=B$).

  3. It is a true statement, with an easy and accessible proof, so presenting it is a very good idea. And it's an excellent opportunity to exercise the definition of what it means for one set to be a subset of another.

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  • $\begingroup$ but X will be the subset of Y. Right? its that a typo or I am getting the concept wrong $\endgroup$ – Shahroz Punjwani Sep 20 '15 at 21:56
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    $\begingroup$ No. If you check the definition of "subset" between $X$ and $Y$ you will see that it does not hold. $\endgroup$ – Asaf Karagila Sep 20 '15 at 21:57
  • $\begingroup$ @ShahrozPunjwani: X is an element of Y, not a subset. Y is {{{}}}; X is {{}}. $\endgroup$ – Deusovi Sep 21 '15 at 3:55
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I wouldn't exactly say that it's important that $\varnothing$ is a subset of every set. Sure, in some contexts it probably is, but that's not really the point.

The point is that from the definition of "subset", we can prove that $\varnothing$ is a subset of all sets; this simply must be so, if we agree on the definitions. So, what's important here is understanding the definition and understanding how to draw conclusions from the definition; using definitions to prove things.

That is what's important here, not necessarily the results themselves (in my opinion, in this specific situation).

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    $\begingroup$ Thank you! :) I think I was looking for this answer. Now I am more clear on it :) $\endgroup$ – Shahroz Punjwani Sep 20 '15 at 21:57
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    $\begingroup$ You're very welcome! Note that there are good reasons why we like to know that $\varnothing$ is a subset of all sets; for one, it makes combinatorial sense, for various counting formulas. It also means we can take advantage of an algebraic structure in situations where we're dealing with sets. Strictly speaking, that's all beyond what's happening here, but it will be nice for the future. $\endgroup$ – pjs36 Sep 20 '15 at 22:01
  • $\begingroup$ :) right! and thank you once again! :) $\endgroup$ – Shahroz Punjwani Sep 20 '15 at 22:02
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The reason for these conventions is to make certain "abstract" statements true in more cases than if there weren't these definitions.

For instance, once can say that $n\leq m$ holds if any only if the set $\{1,\ldots,n\}$ is a subset of $\{1,\ldots,m\}$. There's nothing to argue about if $1\leq n,m$ and also $n\not=m$.

Now what about if $n=0$? Since $0\leq m$ for any natural number $m$, we would want the set $\{1,\ldots,0\}$ to be contained in $\{1,\ldots,m\}$. But what does $\{1,\ldots,0\}$ even mean? We could say that it is the set of all integers $k$ such that $1\leq k\leq 0$, namely the empty set. This would then force us to say that the empty set is a subset of all sets of the form $\{1,\ldots,m\}$, if we wanted our pattern to continue.

Also for the case $n=m$, if we wanted to the pattern to hold we have to say that every set is a subset of itself.

Long story short: We make these definitions so that patterns continue to hold in all cases.

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  • $\begingroup$ I think I am getting at what you are trying to say! It brings alot of clarity. Thank you!!!! $\endgroup$ – Shahroz Punjwani Sep 20 '15 at 22:01
  • $\begingroup$ Read your answer twice and thrice and now I think I understand it $\endgroup$ – Shahroz Punjwani Sep 20 '15 at 22:04
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It's due to axiom of specification which states that to every set A and every condition S(x) there exists a set B whose elements are exactly those elements of A for which S(x) is true. Now if your condition is such that no element of x satisfies that condition,this will create an empty set. And if your condition is such that every element satisfies it, then B=A. from here A and empty set are considered as subsets.

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  • $\begingroup$ I think I am getting what you are trying to say. So in many situations there can actually he Null sets (meaning no elements) which satisfy tht situation. Thank you!! $\endgroup$ – Shahroz Punjwani Sep 20 '15 at 21:59
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$A$ is a subset of $B$ exactly when every element of $A$ is also an element of $B$. Consider what it means, then, for $A$ to not be a subset of $B$; it means precisely that there is some element of $A$ that is not an element of $B$.

Since there are no elements of $\emptyset$, then for any set $C$, there cannot be something which is an element of $\emptyset$ and not an element of $C$, so $\emptyset$ must be a subset of $C$.

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