The question is in the title. It is trivial that $X \cap Y \subseteq X$. Because $X \cap Y$ only contains elements that are both in $X$ and in $Y$. So every element in $X \cap Y$ is also an element of $X$ and so $X \cap Y \subseteq X$ is true.
Is there a more formal way to proof this?
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2$\begingroup$ Your proof is fine and sufficienlty formal. Good job (: $\endgroup$ – Stefan Mesken Jul 25 '16 at 22:42
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Use the definition of subset ($A\subseteq B$ if $x\in A$ implies $x\in B$) and the definition of intersection.
Let $x\in X\cap Y$. Then $x\in X$ (and $x\in Y$) by the definition of intersection.
