# 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 $S$.

### Definition 1

The image of $X$ (under $f$) is defined and denoted as:

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

### Definition 2

The image of $X$ under $f$ is the element of the codomain of the direct image mapping $f^\to: \powerset S \to \powerset T$ of $f$:

$\forall X \in \powerset S: \map {f^\to} X := \set {t \in T: \exists s \in X: \map f s = t}$

Thus:

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

and so the image of $X$ under $f$ is also seen referred to as the direct image of $X$ under $f$.

## Also known as

The term image set is often seen for the image of a subset under a mapping.

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 $f$.

## Notation

In parallel with the notation $f \sqbrk X$ for the direct image mapping of $f$, $\mathsf{Pr} \infty \mathsf{fWiki}$ also employs the notation $\map {f^\to} X$.

This latter notation is used in, for example, T.S. Blyth: Set Theory and Abstract Algebra, and is referred to as the mapping induced by $f$:

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. ... We shall say that the mappings $f^\to$ and $f^\gets$ are the mappings which are induced on the power sets by the mapping $f$.

In a similar manner, the notation $f^{-1} \sqbrk X$, for the premage of a subset under a mapping, otherwise known as the inverse image mapping of $f$, also has the notation $\map {f^\gets} X$ used for it.

Some older sources use the notation $f \mathbin{} X$ for $f \sqbrk X$.

Sources which use the notation $s f$ for $\map f s$ may also use $S f$ or $S^f$ for $f \sqbrk S$.

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:

 $\ds S$ $=$ $\ds \set {0, 1, 2, 3, 4, 5}$ $\ds T$ $=$ $\ds \set {0, 1, 2, 3}$

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

 $\ds f \paren 0$ $=$ $\ds 0$ $\ds f \paren 1$ $=$ $\ds 0$ $\ds f \paren 2$ $=$ $\ds 0$ $\ds f \paren 3$ $=$ $\ds 1$ $\ds f \paren 4$ $=$ $\ds 1$ $\ds f \paren 5$ $=$ $\ds 3$

Let:

 $\ds A$ $=$ $\ds \set {0, 3}$ $\ds B$ $=$ $\ds \set {0, 1, 3}$ $\ds C$ $=$ $\ds \set {0, 1, 2}$

Then:

 $\ds f \sqbrk A$ $=$ $\ds \set {0, 1}$ $\ds f \sqbrk B$ $=$ $\ds \set {0, 1}$ $\ds f \sqbrk C$ $=$ $\ds \set 0$

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:

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

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

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