2024-12-08

896: Map Preimage of Subset Minus Subset Is Preimage of 1st Subset Minus Preimage of 2nd Subset

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description/proof of that map preimage of subset minus subset is preimage of 1st subset minus preimage of 2nd subset

Topics


About: set

The table of contents of this article

Starting Context

Target Context


  • The reader will have a description and a proof of the proposition that for any map, the preimage of any subset minus any subset is the preimage of the 1st subset minus the preimage of the 2nd subset.

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: Structured Description


Here is the rules of Structured Description.

Entities:
S1: ∈{ the sets }
S2: ∈{ the sets }
f: S1→S2
S2,1: ⊆S2
S2,2: ⊆S2
//

Statements:
f−1(S2,1∖S2,2)=f−1(S2,1)∖f−1(S2,2)
//

2: Natural Language Description


For any sets, S1,S2, any map, f:S1→S2, and any subsets, S2,1,S2,2⊆S2, f−1(S2,1∖S2,2)=f−1(S2,1)∖f−1(S2,2).

3: Proof


Whole Strategy: Step 1: see that f−1(S2,1∖S2,2)⊆f−1(S2,1)∖f−1(S2,2); Step 2: see that f−1(S2,1)∖f−1(S2,2)⊆f−1(S2,1∖S2,2).

Step 1:

For each point, p∈f−1(S2,1∖S2,2), f(p)∈S2,1∖S2,2, f(p)∈S2,1 and f(p)∉S2,2, p∈f−1(S2,1) and p∉f−1(S2,2), and so, p∈f−1(S2,1)∖f−1(S2,2).

Step 2:

For each point, p∈f−1(S2,1)∖f−1(S2,2), p∈f−1(S2,1) and p∉f−1(S2,2), f(p)∈S2,1 and f(p)∉S2,2, f(p)∈S2,1∖S2,2, and so, p∈f−1(S2,1∖S2,2).

4: Note


Compare this with the proposition on images.

References
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