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What is the intersection of the sets $\{1\}$ and $\{1,2\}$?

For me, it would make sense that $\{1\} \cap \{1,2\} = 1$, but I'm afraid it must be $\{1\}$, otherwise for instance $T = \{ \{\}, \{1\}, \{1,2\} \}$ would not be a topology of $X = \{1,2\}$ since $1$ does not belong to $T$.

What is the reason that $\{1\} \cap \{1,2\}$ could not be $1$?

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    $\begingroup$ Intersection of sets is a set itself. So, yes, it should be $\{1\}$. $\endgroup$ – Kaster Feb 12 '16 at 0:46
  • $\begingroup$ Your question is sensible, but you should find a better title. "Question" is too uninformative. In any case, yes the intersection should be $\{1\}$. $\endgroup$ – Gregory Grant Feb 12 '16 at 0:48
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    $\begingroup$ Suppose for a moment that $\{1\}\cap\{1,2\}$ is $1$. What's the intersection of $\{1,2,3\}$ and $\{1,2\}$? You would have it be $1,2$, but that's not a thing. $\endgroup$ – Akiva Weinberger Feb 12 '16 at 0:50
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The intersection of two sets is a set. $1$ is an element of the set $\{1\}$.

That is, $X\cap Y=\{x:x\in X\:\text{and}\: x\in Y\}$.

In our case, $X=\{1,2\}, Y=\{1\}$, so $X\cap Y=\{1\}$, the set of elements in both sets $X,Y$.

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Let $X, Y$ be sets. Then $X \cap Y$ is the set whose elements are precisely those that are members of both $X$ and $Y$.

So for $\{1\} \cap \{1,2\}$ this yields a set with precisely one element, namely $1$. Therefore $\{1\} \cap \{1,2\} = \{1\}$.

I don't think it is to address why $1$ isn't the correct answer in a formal way - as this would require to talk about $1$ as a set. Instead let me say the following:

Think of a set as a bag which does contain certain elements. $\{1,2\}$ is a bag and in this bag are precisely two elements - namely the natural numbers $1$ and $2$. Now $\{1\}$ is another bag which only contains $1$ as its only element. If we take their intersection, we build another bag and we put precisely those element in this new bag that appear in both bags. So this bag would contain only one element, namely the natural number $1$. The answer therefore is not $1$, but "the bag that contains $1$ as its only element". Replace "bag" with "set" and you see that $\{1\}$ is the right answer.

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