# How should this definition of a family with corresponding index set be interpreted?

Does the following definition of a family imply that there exist a surjection from $I$ onto the family ? Or should it be interpreted as a injective partial function from $I$ to the family ?

Should I understand it as: some elements of $I$ may not be used to index the elements of the family, but there are enough elements of $I$, such that every element of the family has an unique index ?

The if part "if for each $i \in I$ we have an associated object $x_i$" imply that an index set may have elements not mapped to an element of the family ?

But the bottom part of the definition make it seems that a subset of $\mathbb Z$ neccesarily are mapped to the family in an injective fashion ?

$\quad$ A family is a collection of objects, indexed by some set $I$, called an index set. If for each $i\in I$ we have an associated object $x_i$, the family of all such objects is denoted by $\{x_i\}_{i\in I}$. Unlike a set, a family may contain duplicates; that is, we may have $x_i=x_j$ for some pair of indices $i,j$ with $i\neq j$. If the index set $I$ has some natural order, then we may view the family as being ordered in the same way. As a special case, a family is indexed by a subset of $\mathbb Z$ of the form $\{m,\ldots,n\}$ or $\{m,m+1,\ldots\}$ is a sequence, which we may write as $\{x_i\}^n_{i=m}$ or $\{x_i\}^\infty_{i=m}$.

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So when $A$ is indexed by $I$ not every element of $I$ must correspond to some element of $A$? – user111854 May 27 '14 at 10:32
In my opinion, a family should be interpreted as nothing else but a map (function) $x : I \to X$, defined everywhere on the index set $I$ and taking values in a prescribed set $X$ (which is not explicitly mentioned in the definition given in the question). For instance, an infinite sequence of real numbers is a map $x: \mathbb{N} \to \mathbb{R}$. The 'if' statement that the you refer to explains what it means for a family to be given. It does not imply that the association of any element of the index set with an object is optional (otherwise some elements of the index set would be redundant). – ivanpenev May 27 '14 at 10:39
Regarding surjectivity: Suppose that $x: I \to X$ is a 'family' of elements of $X$ indexed by $I$. Let $Y$ be the set of all $y$ in $X$ such that $y = x_i$ for some $i \in I$. This is the range of the map $x : I \to X$, and by abuse of language, $x$ can also be considered as a map $x: I \to Y$, which by definition would be surjective (onto $Y$). This does not mean that $x$ takes all elements of $X$ as values. For instance, no sequence of real numbers could enumerate all real numbers. – ivanpenev May 27 '14 at 11:07
Regarding injectivity: As stated in the text above, one may have $x_i = x_j$ (as elements of $X$) for some $i \neq j$, which means that $x:I \to X$ need not be injective. On the other hand, one still refers to the $i$th and $j$th members of the family $x$ as distinct members of that family, meaning that their indices are distinct. There is a closely related map $\tilde{x} : I \to I\times X$ defined by $\tilde{x}(i) = (i,x_i)$, which is by construction injective (it's range is by definition the graph of $x$). Hope this doesn't make things more obscure, instead of clarifying them. – ivanpenev May 27 '14 at 11:23
Thank you very much @ivanpenev. – user111854 May 27 '14 at 11:30

You should think of the indexed collection as a "function": for every $i \in I$ we have some well determined set $x_i$. This set $x_i$ can be empty, or we can have, as stated, that for different $i \neq j$ in $I$ we have these et set $x_i = x_j$. But this can also happen for functions (constant functions, or $x \rightarrow x^2$ that have the same image for $x=1$ and $x=-1$). So we always have an $x_i$ for every $i$.
Axiomatic/formal remark: The fact that $\{x_i: i \in I\}$ is a set (when $I$ is a set, and all $x_i$ are, and $x_i$ is "uniquely determined by $i$") is a consequence of the axiom of replacement.
And no, if we use $\mathbb{Z}$ or $\mathbb{N}$ or a finite subset as an index set, this does not mean that the assignment $n$ to $x_n$ is injective. We can easily have a sequence $(1,1,\ldots)$ indexed by $\mathbb{N}$ that is constant. Here $x_n = 1$ for all $n \in \mathbb{N}$, e.g.