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I am reading "Topology 2nd Edition" by James R. Munkres.

The following definition is in this book. (on p.36 section 5)

Definition. Let $\mathcal{A}$ be a nonempty collection of sets. An indexing function for $\mathcal{A}$ is a surjective function $f$ from some set $J$, called the index set, to $\mathcal{A}$. The collection $\mathcal{A}$, together with the indexing function $f$, is called an indexed family of sets. Given $\alpha\in J$, we shall denote the set $f(\alpha)$ by the symbol $A_\alpha$. And we shall denote the indexed family itself by the symbol $$\{A_\alpha\}_{\alpha\in J},$$ which is read "the family of $A_\alpha$, as $\alpha$ ranges over $J$." Sometimes we write merely $\{A_\alpha\}$, if it is clear what the index set is.

Why is an indexing function required to be surjective?

I think we have no trouble without the requirement to be surjective.

José Carlos Santos
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tchappy ha
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    If you want to list your entire collection have to take $f$ surjective. – Kavi Rama Murthy Oct 22 '21 at 08:59
  • @KaviRamaMurthy I think we can require $f$ is surjective only when we want to list our entire collection. – tchappy ha Oct 22 '21 at 09:02
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    @KaviRamaMurthy Let $f$ be not surjective. If we say $f$ is an indexing function for $\mathcal{A}$, I feel it is strange. – tchappy ha Oct 22 '21 at 09:05
  • @KaviRamaMurthy Thank you very much for your comment. – tchappy ha Oct 22 '21 at 09:22
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    If the indexing is injective, it's often called a "faithful" indexing (every set in a set of sets then occurs once and only once in the indexing). This can be handy in some proofs. Set theory does this alot when we enumerate every subset in some well-ordered sequence using cardinals in order to do constructions of strange objects by transfinite recursion (like Bernstein sets). Topologists are also quite fond of it sometimes... – Henno Brandsma Oct 23 '21 at 08:58
  • @HennoBrandsma Thank you very much for your comment. – tchappy ha Oct 23 '21 at 12:26

2 Answers2

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Surjectivity is needed in order that every element of the set $\mathcal A$ is associated to a label, which is something that we want.

José Carlos Santos
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In my opinion it is not really relevant whether one requires $f$ to be surjective or not. As José Carlos Santos has pointed out in his answer, if $f$ is not surjective, then some $A \in \mathcal A$ remain unindexed which we probably do not want.

What does Mukres do with $f$? He writes $A_\alpha$ for $f(\alpha)$ and defines

  • the union $\bigcup_{\alpha \in J} A_\alpha$

  • the intersection $\bigcap_{\alpha \in J} A_\alpha$

  • the Cartesian product $\prod_{\alpha \in J} A_\alpha$

This can be done for any function $f : J \to \mathcal A$, i.e. for any indexed collection of sets $(A_\alpha)_{\alpha \in J}$ in $\mathcal A$.

But why do we need index functions at all? Why doesn't it suffice to work with collections of sets $\mathcal A$? In other words, why don't we simply index $\mathcal A$ by itself (i.e. take $J = \mathcal A$ and $f = id$)? Yes, we can do that, and in this case we get for example

  • the union $\bigcup_{A \in \mathcal A} A = \bigcup \mathcal A$

  • the intersection $\bigcap_{A \in \mathcal A} A = \bigcap \mathcal A$

For an arbitray function $f : J \to \mathcal A$ we get

  • $\bigcup_{\alpha \in J} A_\alpha = \bigcup_{A \in f(J)} A = \bigcup \mathcal f(J)$

  • $\bigcap_{\alpha \in J} A_\alpha = \bigcap_{A \in f(J)} A = \bigcap \mathcal f(J)$

and this shows that the surjectivity of $f$ is a useful property.

Note that we can always replace $\mathcal A$ by the subset $\mathcal A' = f(J) \subset \mathcal A$ and thereby get a surjective index function $f' : J \stackrel{f}{\to} \mathcal A'$ which is essentially the same as $f$.

The main point is that in an indexed collection sets may occur repeatedly (which is impossible in $\mathcal A$ itself). For example, if $\mathcal A =\{A\}$ has only one member, we can nevertheless form arbitrary indexed collections $(A_\alpha)_{\alpha \in J}$ with $A_\alpha = A$ and consider the Cartesian product $\prod_{\alpha \in J} A_\alpha$. This does not agree with $\prod_{A \in \mathcal A} A$ (which is just $A$).

Let us finally observe that in practical applications it may even be even inconvenient to require index functions to be surjective. For example, in a topological space $X$ one often assigns to each point $x \in X$ an open neigborhood $U_x$ of $x$ in $X$. This may be regarded as a function $f : X \to \mathfrak T_X, f(x) = U_x$, where $\mathfrak T_X$ is the topology of $X$. In general such functions will not be surjective and we do not explictly know what $f(X)$ looks like. Nevertheless I tend to accept $f$ as index function.

Paul Frost
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