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Let $A=P(\{1,2,3,4\})$. Let $h$ be the following function.

$$h : \mathbb N \to A$$ defined by $$h(x) = \{2,3\}\cap \{x\}.$$

Write down $h(1)$

I'm a bit puzzled by this question. Does this just mean that $x = 1$, therefore, there is no intersection or is there an actual answer to this question?

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    $\begingroup$ It’s the empty set $\endgroup$ Apr 16, 2019 at 9:27
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    $\begingroup$ Yes, $h(1)$ means that you substitute $x=1$. But there still is an actual answer to the question, and that is $\{\}$. $\endgroup$
    – 5xum
    Apr 16, 2019 at 9:36

2 Answers 2

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$X\cap Y$ means "The set of all elements which are both in $X$ and in $Y$". In the case of $h(1)$, we have $$ h(1) = \{2, 3\}\cap \{1\} $$ There are no elements which are in both $\{2, 3\}$ and in $\{1\}$, so the result of that intersection is $\varnothing = \{\phantom a\}$.

And yes, in this case (as in most concrete cases), $h(1)$ does mean "Take any place in the definition of $h(x)$ where $x$ appears, exchange it with a $1$, and then calculate."

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Hint

$A = \mathcal P( \{ 1,2,3,4 \})$ is the power-set of set $A$,

i.e. the set of subsets of $A$.

$h : \mathbb N → A$, defined by $h(x) = \{ 2,3 \} \cap \{ x \}$,

is a function that assign to each natural number $n$ a set: the intersection of the set $\{ 2,3 \}$ with the singleton set $\{ x \}$.

Thus, we may perform some simple checks : with $x=2$ we have that $h(2) = \{ 2,3 \} \cap \{ 2 \} = \{ 2 \}$.

Now, what about $x=1$ ?

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