Image of Set Difference under Injection

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $S_1$ and $S_2$ be subsets of $S$.

Let $S_1 \setminus S_2$ denote the set difference between $S_1$ and $S_2$.


Then:

$\forall S_1, S_2 \subseteq S: f \sqbrk {S_1} \setminus f \sqbrk {S_2} = f \sqbrk {S_1 \setminus S_2}$

if and only if $f$ is an injection.


Proof

An injection is a type of one-to-one relation, and therefore also a one-to-many relation.


Therefore One-to-Many Image of Set Difference applies:

$\RR \sqbrk {S_1} \setminus \RR \sqbrk {S_2} = \RR \sqbrk {S_1 \setminus S_2}$

if and only if $\RR$ is one-to-many.


We have that $f$ is a mapping and therefore a many-to-one relation.

So $f$ is a one-to-many relation if and only if $f$ is also an injection.


It follows that:

$\forall S_1, S_2 \subseteq S: f \sqbrk {S_1} \setminus f \sqbrk {S_2} = f \sqbrk {S_1 \setminus S_2}$

if and only if $f$ is an injection.

$\blacksquare$


Sources


This page may be the result of a refactoring operation.
As such, the following source works, along with any process flow, will need to be reviewed.
When this has been completed, the citation of that source work (if it is appropriate that it stay on this page) is to be placed above this message, into the usual chronological ordering.
In particular: extract Halmos' result into a different page. It looks like it can be formulated in terms of complement
If you have access to any of these works, then you are invited to review this list, and make any necessary corrections.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{SourceReview}} from the code.


Retrieved from "https://proofwiki.org/w/index.php?title=Image_of_Set_Difference_under_Injection&oldid=545749"