Mapping Image of Intersection

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Theorem

The image of the intersection of subsets of a mapping is a subset of the intersection of their images.


That is:

Let $f: S \to T$ be a mapping. Let $S_1$ and $S_2$ be subsets of $S$.


Then:

$f \left({S_1 \cap S_2}\right) \subseteq f \left({S_1}\right) \cap f \left({S_2}\right)$


General Result

Let $f: S \to T$ be a mapping.

Let $\mathcal P \left({S}\right)$ be the power set of $S$.

Let $\mathbb S \subseteq \mathcal P \left({S}\right)$.


Then:

$\displaystyle f \left({\bigcap \mathbb S}\right) \subseteq \bigcap_{X \in \mathbb S} f \left({X}\right)$


Proof

As $f$, being a mapping, is also a relation, we can apply Image of Intersection:

$\mathcal R \left({S_1 \cap S_2}\right) \subseteq \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$

and

$\displaystyle \mathcal R \left({\bigcap \mathbb S}\right) \subseteq \bigcap_{X \in \mathbb S} \mathcal R \left({X}\right)$

$\blacksquare$


Note

Note that equality does not hold in general.

Let:

  • $S_1 = \left\{{x \in \Z: x \le 0}\right\}$
  • $S_2 = \left\{{x \in \Z: x \ge 0}\right\}$
  • $f: \Z \to \Z: \forall x \in \Z: f \left({x}\right) = x^2$

We have:

  • $f \left({S_1}\right) = \left\{{0, 1, 4, 9, 16, \ldots}\right\} = f \left({S_2}\right)$

Then:

$f \left({S_1}\right) \cap f \left({S_2}\right) = \left\{{0, 1, 4, 9, 16, \ldots}\right\}$

but:

$f \left({S_1 \cap S_2}\right) = f \left({\left\{{0}\right\}}\right) = \left\{{0}\right\}$


Note that from Injection Image of Intersections equality always holds iff $f$ is an injection.


Also see



Sources

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