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Here is the context for my question:

Let A = {1,2,5,8,11}.

Here is my question:

Is ∅ ⊆ A?

Why or why not?

Asaf Karagila
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Sage Hopkins
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  • I took a few minutes to contemplate whether or not this is a duplicate, or should be closed for lack of effort. I chose the former, but you should search rudimentary questions before posting them, and you should definitely include your own thoughts and were the question had you stumped in the future. – Asaf Karagila Jan 28 '16 at 21:51
  • I'm not sure if I should say this, but it might help intuitively. A smaller set is a subset of a larger set if you can get from the big set to the little set by removing items. You can always get to the empty set by removing everything. So the empty set is a subset of every set. I don't know if I should have put it like that as it abuses math's precision and isn't entirely accurate but it does avoid the emp set not in a set so how can it be a subset issue, that is so hard for some people. – fleablood Jan 28 '16 at 22:04

1 Answers1

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The empty set is a subset of every set. The subset relation $\subseteq$ is defined as a shorthand: $X\subseteq Y$ means $\forall x(x\in X\to x\in Y)$. Thus, $\emptyset\subseteq X$ means $\forall x(x\in\emptyset\to x\in X)$. Because $x\in\emptyset$ is never true, the entire inner formula $(x\in\emptyset\to x\in X)$ is always true, "vacuously" — the value of the conditional $p\to q$ is true when $p$ is false — so the inner formula is true for all $x$.

In particular, $\emptyset$ is a subset of your $A$.

BrianO
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  • I'm not sure I would use the word "contained," as it sounds like $\in$ to me, rather than $\subseteq$. (Though I see that's the word OP used.) – Akiva Weinberger Jan 28 '16 at 21:45
  • @AkivaWeinberger Yyyyeah I was just following OP's usage... with some unease of my own. I changed it: now we can both relax a little. – BrianO Jan 28 '16 at 21:46
  • You meant $\forall x(x \in \color{red}{X} \to x \in Y)$. – N. F. Taussig Jan 28 '16 at 21:53
  • @N.F.Taussig Sure do — fixed. Thanks. – BrianO Jan 28 '16 at 21:56
  • I think "contained" demonstrates confusion and ambiguity. This is semantically confusing and very, common misunderstanding. A set can be said to contain elements, but it can also be said to "contain" a subset. Best to avoid the term altogether – fleablood Jan 28 '16 at 21:57
  • @fleablood Agreed. There is attested usage: $\subseteq$ has been referred to as "containment". But in other usage, "$1$ is contained in ${1}$". So, best to pass on the potential confusion, especially for beginners. – BrianO Jan 28 '16 at 22:00