294: Map Image of Intersection of Sets Is Not Necessarily Intersection of Map Images of Sets

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that map image of intersection of sets is not necessarily intersection of map images of sets

---

Topics

About: set
About: map

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

---

Starting Context

• The reader knows a definition of map.

---

Target Context

• The reader will have a description and a proof of the proposition that for a map, the map image of an intersection of sets is not necessarily the intersection of the map images of the sets.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Description

For some sets, $S_1$ and $S_2$, a map, $f:S_1\to S_2$, and a possibly uncountable number of subsets of $S_1$, $S_{1_i}\subseteq S_1$, the map image of the intersection of the subsets, $f({\cap }_i{S}_{1_i})$, is not necessarily the intersection of the map images of the subsets, ${\cap }_{i}f(S_{1_i})$, which is, not necessarily $f({\cap }_i{S}_{1_i})={\cap }_{i}f(S_{1_i})$.

---

2: Proof

It suffices to show a counter-example. If each element in $S_1$ maps to the same element, $p\in S_2$, $S_{1_1}$ is a non-empty proper subset, $S_{1_1}\subset S_1$, and $S_{1_2}$ is the complement of $S_{1_1}$ with respect to $S_1$, $S_{1_2}=S_1\setminus S_{1_1}$, ${\cap }_{i}f(S_{1_i})=p$, but ${\cap }_i{S}_{1_i}=\mathrm{\varnothing }$, so $f({\cap }_i{S}_{1_i})=\mathrm{\varnothing }$.

---

3: Note

Although $f({\cup }_i{S}_{1_i})={\cup }_{i}f(S_{1_i})$ always holds, the intersection counterpart does not necessarily holds.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>