Definition:Image (Set Theory)/Mapping/Subset

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Definition

Let $S$ and $T$ be sets.

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

Let $X \subseteq S$ be a subset of $S$.

Then the image of $X$ (under $f$) is defined as:

$f \sqbrk X := \set {t \in T: \exists s \in X: f \paren s = t}$


That is:

$f \sqbrk X := \map {f^\to} X$

where $f^\to$ denotes the mapping induced on the power set of $S$ by $f$.


Also known as

The term image set is often seen for image.

The modifier by $f$ can also be used for under $f$.

Thus, for example, the image set of $X$ by $f$ means the same as the image of $X$ under $f$.


Some authors stipulate the name further by calling $f \sqbrk X$ the direct image of $X$ (by $f$).

This is done to emphasize the distinction between this and the concept of the inverse image.


Notation

As well as using the notation $\Img f$ to denote the image set of a mapping, the symbol $\operatorname {Img}$ can also be used as follows:

$\operatorname {Img}_f \paren X := f \sqbrk X$

but this notation is rarely seen.


Similarly obscure is the notation $f \mathbin{``} X$ for $f \sqbrk X$, which is mainly encountered in older accounts on set theory.


Some authors prefer not to use the notation $f \sqbrk X$ and instead use the concept of the mapping induced from the power set of $S$ to the power set of $T$.

For example, T.S. Blyth: Set Theory and Abstract Algebra uses $\map {f^\to} X$ for $f \sqbrk X$:

It should be noted that most mathematicians write $\map f X$ for $\map {f^\to} X$. Now it is quite clear that the mappings $f$ and $f^\to$ are not the same, so we shall retain the notation $f^\to$ to avoid confusion.


Some authors do not bother to make the distinction between the image of an element and the image set of a subset, and use the same notation for both:

The notation is bad but not catastrophic. What is bad about it is that if $A$ happens to be both an element of $X$ and a subset of $X$ (an unlikely situation, but far from an impossible one), then the symbol $\map f A$ is ambiguous. Does it mean the value of $f$ at $A$ or does it mean the set of values of $f$ at the elements of $A$? Following normal mathematical custom, we shall use the bad notation, relying on context, and, on the rare occasions when it is necessary, adding verbal stipulations, to avoid confusion.
-- 1960: Paul R. Halmos: Naive Set Theory

Similarly, Allan Clark: Elements of Abstract Algebra, which uses the notation $f x$ for what $\mathsf{Pr} \infty \mathsf{fWiki}$ denotes as $\map f x$, also uses $f X$ for $f \sqbrk X$ without comment on the implications.

In the same way does John D. Dixon: Problems in Group Theory provide us with $S^f$ for $f \sqbrk S$ as an alternative to $\map f S$, again making no notational distinction between the image of the subset and the image of the element.

On $\mathsf{Pr} \infty \mathsf{fWiki}$ this point of view is not endorsed.


Some authors recognise the confusion, and call attention to it, but don't actually do anything about it:

In this way we obtain a map from the set $\powerset X$ of subsets of $X$ to $\powerset Y$; this map is still denoted by $f$, although strictly speaking it should be given a different name.
-- 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra


Examples

Aribtrary Mapping from $\set {0, 1, 2, 3, 4, 5}$ to $\set {0, 1, 2, 3}$

Let:

\(\displaystyle S\) \(=\) \(\displaystyle \set {0, 1, 2, 3, 4, 5}\) $\quad$ $\quad$
\(\displaystyle T\) \(=\) \(\displaystyle \set {0, 1, 2, 3}\) $\quad$ $\quad$


Let $f: S \to S$ be the mapping defined as:

\(\displaystyle f \paren 0\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle f \paren 1\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle f \paren 2\) \(=\) \(\displaystyle 0\) $\quad$ $\quad$
\(\displaystyle f \paren 3\) \(=\) \(\displaystyle 1\) $\quad$ $\quad$
\(\displaystyle f \paren 4\) \(=\) \(\displaystyle 1\) $\quad$ $\quad$
\(\displaystyle f \paren 5\) \(=\) \(\displaystyle 3\) $\quad$ $\quad$


Let:

\(\displaystyle A\) \(=\) \(\displaystyle \set {0, 3}\) $\quad$ $\quad$
\(\displaystyle B\) \(=\) \(\displaystyle \set {0, 1, 3}\) $\quad$ $\quad$
\(\displaystyle C\) \(=\) \(\displaystyle \set {0, 1, 2}\) $\quad$ $\quad$


Then:

\(\displaystyle f \sqbrk A\) \(=\) \(\displaystyle \set {0, 1}\) $\quad$ $\quad$
\(\displaystyle f \sqbrk B\) \(=\) \(\displaystyle \set {0, 1}\) $\quad$ $\quad$
\(\displaystyle f \sqbrk C\) \(=\) \(\displaystyle \set 0\) $\quad$ $\quad$


and:

$\Img f = \set {0, 1, 3}$


Image of $\closedint {-3} 2$ under $x \mapsto x^4 - 1$

Let $f: \R \to \R$ be the mapping defined as:

$\forall x \in \R: f \paren x = x^4 - 1$

The image of the closed interval $\closedint {-3} 2$ is:

$f \sqbrk {\paren {\closedint {-3} 2} } = \closedint {-1} {80}$


Also see


Sources