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If $S$ is a nonempty set, is the following statement correct: $\varnothing \subset S$? It's confusing me becuase $\varnothing$ does not contain any elements so I'm struggling with the logic behind this statement.

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Yes, it's correct. To say "$X \subseteq Y$" means that everything which is an element of $X$ is also an element of $Y$. Now imagine how you would demonstrate this wrong: you would supply an element of $X$ that is not an element of $Y$.

So, to show that $\emptyset \subseteq S$ is false, we would have to supply an element of $\emptyset$ that is not an element of $S$. Since we can't supply an element of $\emptyset$ at all, no such counterexample can be found; so the statement is true.

Now, the symbol $\subset$ denotes proper subset - $X \subset Y$ if and only if $X \subseteq Y$ and $X \neq Y$. So $\emptyset \subset S$ if and only if $\emptyset \subseteq S$ (which is always true) and $\emptyset \neq S$ (which is true if $S$ is nonempty).

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  • $\begingroup$ So $A\subset B$ should be interpreted as "there are no $x\in A$ such that $x \notin B$ instead of "all $x \in A$ also satisfy $x \in B$" since there might be no $x \in A$? $\endgroup$ – David Nov 25 '16 at 14:32
  • $\begingroup$ @David By convention, the two statements are considered equivalent; but the first might be more intuitive. $\endgroup$ – Reese Nov 25 '16 at 14:35
  • $\begingroup$ Indeed it is, thanks a lot. $\endgroup$ – David Nov 25 '16 at 14:35
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The statement is vacuously true: suppose the proposition is false: therefore $\emptyset \not\subset S$. Therefore, there must exist $s\in \emptyset$ such that $s \not\in S$. But $\emptyset$ does not contain any element, meaning $s$ cannot exist. It must be true because stating otherwise is impossible.

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  • $\begingroup$ Does one always have to consider a statement to be either true or false? Can't it be undefined under certain circumstances? $\endgroup$ – David Nov 25 '16 at 14:41
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$\varnothing\subset S$ for any set $S$, since you cannot find a counter-example, even if $S=\varnothing$: an element in $\varnothing$ which would not be in $S$.

Note: The correct mathematical notation is $\varnothing$ (code \varnothing), not $\emptyset$.

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    $\begingroup$ -1 for saying that $\varnothing$ is correct, but $\emptyset$ isn't. $\endgroup$ – Magdiragdag Nov 25 '16 at 14:23
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    $\begingroup$ From context, it seems that OP is distinguishing between $\subset$ and $\subseteq$; so $\subset$ means strict subset. $\endgroup$ – Reese Nov 25 '16 at 14:24
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    $\begingroup$ You do as you please, but it's more of a computer science notation. I stick to Bourbaki's notation. $\endgroup$ – Bernard Nov 25 '16 at 14:24
  • $\begingroup$ @Reese: Well, strict is denoted $\subsetneq$ or $\varsubsetneq$. $\endgroup$ – Bernard Nov 25 '16 at 14:26
  • $\begingroup$ Why do you say $\emptyset$ is not correct? It is written \emptyset. The meaning is clear. +1 because the answer is correct. $\endgroup$ – mfl Nov 25 '16 at 14:27
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It is true because it is not false that every object in $\emptyset$ belongs to $S$.

Let me elaborate.

$x \in \emptyset$ but $\emptyset$ is empty so whatever you want to say about any object inside $\emptyset$ will always be true since none exist, you might as well say that $x \in \emptyset$.

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