$\begingroup$ "Subset of" method something different than "element of". Keep in mind $\a\$ is additionally a subset of $X$, despite $\ a \$ not appearing "in" $X$. $\endgroup$

Because every single element that $\emptyset$ is additionally an facet of $X$. Or have the right to you name an facet of $\emptyset$ that is not an element of $X$?

that"s since there space statements that space vacuously true. $Y\subseteq X$ way for all $y\in Y$, we have $y\in X$. Currently is it true the for every $y\in \emptyset $, we have actually $y\in X$? Yes, the statement is vacuously true, because you can"t pick any $y\in\emptyset$.

You are watching: Empty set is subset of every set

You must start from the an interpretation :

$Y \subseteq X$ iff $\forall x (x \in Y \rightarrow x \in X)$.

Then girlfriend "check" this definition with $\emptyset$ in ar of $Y$ :

$\emptyset \subseteq X$ iff $\forall x (x \in \emptyset \rightarrow x \in X)$.

Now you must use the truth-table definition of $\rightarrow$ ; you have actually that :

"if $p$ is *false*, climate $p \rightarrow q$ is *true*", for $q$ whatever;

so, because of the fact that :

$x \in \emptyset$

is **not** *true*, because that every $x$, the over truth-definition that $\rightarrow$ offers us the :

"for every $x$, $x \in \emptyset \rightarrow x \in X$ is *true*", for $X$ whatever.

This is the factor why the *emptyset* ($\emptyset$) is a *subset* the every set $X$.

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edited Jun 25 "19 in ~ 13:51

answered jan 29 "14 at 21:55

Mauro ALLEGRANZAMauro ALLEGRANZA

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$\endgroup$

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include a comment |

4

$\begingroup$

Subsets room not have to elements. The aspects of $\a,b\$ room $a$ and also $b$. But $\in$ and also $\subseteq$ are different things.

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answered jan 29 "14 in ~ 19:04

Asaf Karagila♦Asaf Karagila

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