# Proof that product topology of subspace is same as induced product topology

Let's assume that $A\subseteq X$ is product of $A_{i}\subseteq X_{i}(i\in I)$.

Then product topology of $A$ is the same than the topology induced by $X$.

I have proved this few different times now, and for this one I need help. I like to try all kinds of proofs.

Collection $\mathcal{B}_{A}$ has sets $V:=\prod_{i\in I} V_i$, where $V_i\subseteq A_{i}$ by every $i\in I$ and $V_i\neq A_i$ finitely many $i$.

Collection $\mathcal{B}_{X}$ has sets $\prod_{i}U_{i\in I}\cap \prod_{i}A_{i\in I}$, where and $U_{i}$ is element of basis of $X_{i}$. Notation $U:=\prod_{i}U_{i\in I}$.

There is theorem that says $\mathcal{B}$ is basis for topology iff every open $U\subseteq X$ can be shown in the form $$U=\bigcup_{a\in A} B_{a},$$ where $B_{a}\in\mathcal{B}$ for all $a\in A$.

Now what I am trying to do in this proof attempt is that I want to try out the theorem above.

There are problems with the notion and that is one main thing where I need tips.

I hope that you get the idea what I am after here.

Proof:

$Z$ belongs in the product topology of $A$.

$\Leftrightarrow$ $$Z=\bigcup_{i\in I} \big(\prod_{i\in I} V_{i} \big)_{i}\quad\text{Where } V_{i}\in \mathcal{T}_{A_{i}}\text{ for all }i\in I.$$ $\Leftrightarrow$ $$Z=\bigcup_{i\in I} \big(\prod_{i \in I}U_{i}\cap A_{i} \big)\quad\text{Where }U_{i}\in\mathcal{T_{i}}\text{ for all }i\in I.$$ $\Leftrightarrow$ $$Z=\bigcup_{i\in i} (\prod_{i\in I} U_{i}\cap \prod_{i\in I} A_{i})_{i}$$ $\Leftrightarrow$ $$Z=\big(\bigcup_{i\in I}\big(\prod_{i\in I} U_{i} \big)_{i} \big)\cap \prod_{i\in I} A_{i}$$ $\Leftrightarrow$

$Z$ belongs to product topology induced by $X$.

• See my post at.yorku.ca/cgi-bin/…, this follows from the very general transitive law for initial topologies. – Henno Brandsma Nov 27 '15 at 11:00
• You should maybe give the $U_i$ an index $a$, since they depend on $a$. Also you could spell out things like $\prod A_i =A$ and $V_a= \prod U_i \cap A_i$. – user60589 Nov 27 '15 at 15:04
• The first and last equivalences are also not optimal in my point of view. You have made a statement which is equivalent to a formula. Maybe you can add some text to the formula to make it the "full" statement. Last but not least leave out the intermediate "$\Leftrightarrow$" in favor of the $"="$. (Disclaimer: what I wrote is maybe personal taste) – user60589 Nov 27 '15 at 15:12

Let's start with a definition: Let $(X,\mathcal{T})$ be a topological space. Let $I$ be an index set, and let $Y_i (i \in I)$ be topological spaces and let $f_i: X \rightarrow Y_i$ be a family of functions. Then $\mathcal{T}$ is called the initial topology with respect to the maps $f_i$ iff

1. $\mathcal{T}$ makes all $f_i$ continuous.
2. If $\mathcal{T}'$ is any other topology on $X$ that makes all $f_i$ continuous, then $\mathcal{T} \subseteq \mathcal{T}'$.

Or put more shortly: $\mathcal{T}$ is the smallest (coarsest) topology that makes all $f_i$ continuous.

Remark: it is not very useful to ask for the largest topology to make all $f_i$ continuous, then we would always get the discrete topology on $X$. This is the largest topology on $X$ and it makes any map continuous.

This is in fact a common way to construct a topology on a set $X$, based on functions to spaces $Y_i$ that already have a topology. E.g. in linear space theory we can consider an alternative topology on a linear space $X$ as the smallest topology that makes all linear maps (in the original topology) from $X$ to $\mathbb{R}$ continuous. This is called the weak topology on the linear space $X$ (as it generally weaker (fewer open sets) than the original topology on $X$.) That this works is based on the following simple:

### Existence theorem for initial topologies:

Let $X$ be a set and $f_i : X \rightarrow Y_i$ be a collection of topological spaces and maps. Then there is a topology on $X$ that is initial w.r.t. the maps $f_i$. Moreover, this topology is unique and a subbase of the topology is given by $\mathcal{S} = \{(f_i)^{-1}[O]: i \in I, O \text{ open in } Y_i\}$.

(a subbase is a collection $\mathcal{S}$ of subsets of $X$ such that all finite intersections of elements of $\mathcal{S}$ form a base for the topology).

Proof: Let $\mathcal{T}$ be the topology generated by $\mathcal{S}$ as a subbase. This means that $\mathcal{T}$ is the collection of all sets that can be written as unions of finite intersections from $\mathcal{S}$. Then all $f_i$ are continuous, as for all open $O \subseteq Y_i$ the inverse image under $f_i$ of $O$ is in $\mathcal{T}$.

And if $\mathcal{T}'$ is any topology that makes all $f_i$ continuous, $\mathcal{T}'$ must contain all sets of the form $(f_i)^{-1}[O]$ and so $\mathcal{T}'$ must contain $\mathcal{S}$, and as $\mathcal{T}'$ is closed under finite intersections and unions, we have that $\mathcal{T} \subseteq \mathcal{T}'$, as required.

The unicity is clear, because if $\mathcal{T}$ and $\mathcal{T}'$ are both initial then $\mathcal{T} \subseteq \mathcal{T}'$ (by 2) applied to $\mathcal{T}$) and $\mathcal{T}' \subseteq \mathcal{T}$ (by 2) applied to $\mathcal{T}'$) and thus $\mathcal{T} = \mathcal{T}'$.

Example 1: if $A$ is a subset of a topological space $X$, and $i: A \rightarrow X$ is the inclusion map from $A$ to $X$ (defined by $i(x) = x$ for all $x \in A$), then the subspace topology on $A$ is just the initial topology w.r.t. $i$.

Proof: the subspace topology is defined by $\{O \cap A: O \text{ open in } X\}$. But $i^{-1}[O] = O \cap A$ ($x \in O$ and $x \in A$, then $i(x) = x$ is in $O$, so $x \in i^{-1}[O]$ and if $x \in i^{-1}[O]$ then $x \in A$ by definition and $x = i(x)$ must be in $O$, so $x$ is in $O \cap A$). We see that the subbase $\mathcal{S}$ from the existence theorem is just equal to the subspace topology!

Remark: in general, when we have just one function $f: X \rightarrow Y$, and $X$ has the initial topology w.r.t. $f$, we get the topology and not just a subbase. Also, if $f$ is moreover injective (one-to-one), then $f$ is called an embedding.

Example 2: If $X_i, i \in I$ is a family of topological spaces and $X$ is the Cartesian product of the spaces $X_i$, then we have the projection maps $p_j: X \rightarrow X_j$ defined by $p_j( (x_i)_{i \in I}) = x_j$ for all $j \in I$. Then the product topology on $X$ is just the initial topology w.r.t. the maps $p_i$ ($i \in I$).

Proof: the sets $(p_j)^{-1}[O]$ are just the product sets of the form $\prod_i O_i$ where all $O_i = X_i$ except $O_j = O$. So the finite intersections of the subbase elements are exactly all such sets $\prod_i O_i$ where finitely many $O_i$ are some open set in their respective coordinate space and in all other coordinates $O_i$ are equal to $X_i$; this is precisely the standard base for the product topology.

So two very common ways of making new spaces from old ones, subspaces and products, are special cases of initial topologies. In the following I will develop some basic theory that will allow us to formulate and prove general principles that will apply to all examples of initial topologies. Some well-known facts can then be seen together in a common framework.

The fact that a space $X$ has the initial topology w.r.t. a family of mappings, makes it easy to recognise continuous functions to $X$. We have the following useful:

### Universal theorem of continuity for initial topologies.

Let $X$ be a space and $f_i : X \rightarrow Y_i$ $(i \in I)$ a family of mappings and spaces $Y_i$, such that $X$ has the initial topology with respect to the $f_i$. Let $Y$ be any space and $g$ a function from $Y$ to $X$. Then $g$ is continuous iff for all $i \in I$: $f_i \circ g$ is continuous from $Y$ to $Y_i$.

Proof: if $g$ is continuous then all $f_i \circ g$ are also continuous, because all $f_i$ are continuous and compositions of continuous maps are continuous. Suppose now that $f_i \circ g$ is continuous for all $i$. Let $S$ be any element of the subbase $\mathcal{S}$ (from the existence theorem), so that $S = (f_i)^{-1}[O]$ for some open subset $O$ of $Y_i$. Now $$g^{-1}[A] = g^{-1}[(f_i)^{-1}[O]] = (f_i \circ g)^{-1}[O]$$ which is open because $f_i \circ g$ is continuous.

This shows that inverse images of elements from $\mathcal{S}$ are open. But then as $g^{-1}$ preserves (finite) intersections and unions and as all open subsets of $X$ are unions of finite intersections of elements from $\mathcal{S}$, we see that $g^{-1}[O]$ is open for all open subsets $O$ of $X$. Or: $g$ is continuous.

There is a converse to this as well:

### Characterisation of the initial topology by the continuity theorem.

Let $X$ be a space, and $f_i: X \rightarrow Y_i$ be a family of spaces and functions. Suppose that $X$ satisfies the universal continuity theorem in the following sense:

(*) for all spaces $Z$, for any function $g: Z \rightarrow X$: $(f_i \circ g)$ is continuous for all $i$ iff $g$ is continuous.

Then $X$ has the initial topology w.r.t. the maps $f_i$.

Proof: the identity on X is always continuous, so applying ($\ast$) from right to left with $g = \operatorname{id}$ gives us that all $f_i$ are continuous. If $\mathcal{T}'$ is another topology on $X$ that makes all $f_i$ continuous, then consider the map $g: (X, \mathcal{T}') \rightarrow (X, \mathcal{T})$, defined by $g(x) = x$. Then all maps $f_i \circ g$ are just the maps $f_i$ as seen between $(X, \mathcal{T}')$ and $Y_i$ which are by assumption continuous. So by the other direction of (*) we see that $g$ is continuous, and thus (as $g(x) = x$, and thus $g^{-1}[O] = O$ for all $O$) we have that $\mathcal{T} \subseteq \mathcal{T}'$, as required for the second property of the initial topology.

Applications: Characterisation of continuity into products: a map $f$ into a product $\prod_{i \in I} X_i$ is continuous iff $p_i \circ f$ is continuous for all $i \in I$.

Or suppose that $X$ is any space, $Y'$ a subspace of a space $Y$, and $g: X \rightarrow Y$ is a map such that $g[X] \subseteq Y'$. Then there is the "image restriction" $g': X \rightarrow Y'$ of $g$, defined by $g'(x) = g(x)$.

Note that, if $i: Y' \rightarrow Y$ is the inclusion, then $Y'$ has the initial topology w.r.t. i, and moreover, $g = g' \circ i$. So the theorem we just proved says: $g$ continuous iff $g'$ continuous. This has the intuitive meaning that continuity of $g$ is only determined by $Y'$. So if $g$ is an embedding (see above), then $g$ is a continuous bijection between $X$ and $g[X]$, which is also open as a map between these spaces, when we give $g[X]$ the subspace topology: let $O$ be open in $X$. Then $O = g^{-1}[O']$ for some open subset of $Y$ (by embedding = initial map), and then $O' \cap g[X]$ is open in g[X], and $g[O] = O' \cap g[X]$, by injectivity of $g$. The reverse is also quite easy to see (exercise): if $g:X \rightarrow g[X] \subseteq Y$ is a homeomorphism, then $g$ is an embedding from $X$ into Y. Many books actually define embeddings that way.

Note that the restriction of $f$ to $A$, $f | A$, is just $f \circ i$, where $i$ is the embedding of $A$ into $X$, so that $f | A$ is continuous as a composition of continuous maps.

Application: diagonal product map.

Let $X$ be a space and let $Y_i$ ($i \in I$) be a family of spaces, and $f_i : X \rightarrow Y_i$ be a family of functions. Let $Y$ be the product of the $Y_i$, with projections $p_i$. Define $f:X \rightarrow Y$, the so-called diagonal product of the $f_i$, as follows: $f(x) = (f_i(x))_{i \in I}$. Then $f$ is continuous iff for all $i \in I$, $f_i$ is continuous.

Proof: immediate from the universal continuity theorem, because for all $i$ we have that $p_i \circ f = f_i$.

Application: product maps.

Let $f_i : X_i \rightarrow Y_i$ be a family of functions between spaces $X_i$ and $Y_i$, let $X = \prod_{i \in I} X_i$, $Y = \prod_{i \in I} Y_i$, and let $f:X \rightarrow Y$ be defined by $f((x_i)_i) = (f_i(x_i))_i$, which is called the product map of the $f_i$.

Then $f$ is continuous iff for all $i \in I$ we have that $f_i$ is continuous.

Proof: let $p_i$ be the projections from $Y$ to $Y_i$, and let $q_i$ be the projections from $X$ to the $X_i$. Then for all $i$ we have $$p_i \circ f = f_i \circ q_i\text{.}$$

Suppose that all $f_i$ are continuous. Then, as all $q_i$ and $f_i$ are continuous, all maps $p_i \circ f$ are continuous. As $Y$ has the initial topology w.r.t. the $p_i$, we have by the universal continuity theorem that $f$ is continuous.

Now let $f$ be continuous. Fix $i$ in $I$. Also take a point $r= (r_i)_{i \in I}$ from $X$. Let $(s_i)_{i \in I}$ in $Y$ be its image $f(r)$.

Then the map $k_i: X_i \rightarrow X$, defined as the diagonal product of the identity on $X_i$ and all constant maps onto the point $r_j$ for all $j \neq i$. By the previous application, this is continuous. Moreover, $q_i \circ k_i$ is the identity on $X_i$, also denoted by $\operatorname{id}_{X_i}$. But note that $$f_i = f_i \circ \operatorname{id}_{X_i} = f_i \circ (q_i \circ k_i) = (f_i \circ q_i) \circ k_i = (p_i \circ f) \circ k_i\text{,}$$ which is continuous, as $f$, $p_i$ and $k_i$ are. So $f_i$ is continuous, for all $i \in I$.

A very useful general fact is the following:

### Transitive law of initial topologies.

Suppose that we have a family of spaces and maps $f_i : X \rightarrow Y_i$ ( $i \in I$) and for each $i \in I$ an index set $I_i$, and a family of maps $g_{i,j} : Y_i \rightarrow Z_j$ for $j$ in $I_i$. Assume that each $Y_i$ has the initial topology w.r.t. the $g_{i,j}$ ($j \in I_i$). Then $X$ has the initial topology w.r.t. the maps $g_{i,j} \circ f_i$ ($i \in I, j \in I_i$) iff $X$ has the initial topology w.r.t. the $f_i$ ($i \in I$).

Proof: Suppose that $X$ has the initial topology w.r.t. the $f_i$. Call this topology $\mathcal{T}$. All $g_{i,j}$ are continuous (part of being initial of the topology on $Y_i$) so all $g_{i,j} \circ f_i$ are continuous. Suppose that $\mathcal{T}'$ is another topology on $X$ that makes all $g_{i,j} \circ f_i$ continuous.

Then consider the maps $f'_i : (X,\mathcal{T}') \rightarrow Y_i$, defined by $f'_i(x) = f_i(x)$. So we have $g_{i,j} \circ f'_i = g_{i,j} \circ f_i$ for all relevant indices.

By assumption all $g_{i,j} \circ f'_i = g_{i,j} \circ f_i : (X,\mathcal{T}') \rightarrow Z_j$ are continuous, and as all $Y_i$ have the initial topology w.r.t. the $g_{i,j}$, we see that all $f'_i$ are (by the universal continuity theorem) continuous. If $\operatorname{id}$ is the identity map from $(X,\mathcal{T}') \rightarrow (X,\mathcal{T})$, then $f_i \circ \operatorname{id} = f'_i$ for all $i$. We have just seen that all $f'_i$ are continuous, and as $\mathcal{T}$ is initial w.r.t. the $f_i$, we see that $\operatorname{id}$ is a continuous map by this same universal continuity theorem. But the identity from $(X,\mathcal{T}') \rightarrow (X,\mathcal{T})$ is continuous iff $\mathcal{T} \subset \mathcal{T}'$ (as $O \in \mathcal{T}$ means $\operatorname{id}^{-1}[O] = O \in \mathcal{T}'$), so $\mathcal{T}$ is indeed minimal w.r.t. the continuity of all maps $g_{i,j} \circ f_i$, and $X$ has the initial topology w.r.t. these maps.

Suppose on the other hand that $X$ has the topology $\mathcal{T}$, which is initial w.r.t. the maps $g_{i,j} \circ f_i$. Let $i$ be in $I$. For all $j \in I_i$ we know that $g_{i,j} \circ f_i$ is continuous. As $Y_i$ has the initial topology w.r.t. the maps $g_{i,j}$ ($j \in I_i$), we see again by the universal continuity theorem that $f_i$ is continuous. So all $f_i$ (from $(X,\mathcal{T})$ to $Y_i$) are continuous. Let $\mathcal{T}'$ be another topology on $X$ that makes all $f_i$ continuous. This means that all $g_{i,j} \circ f_i$ are continuous, and so by minimality of $\mathcal{T}$ (by the definition of initial topology w.r.t. the maps $g_{i,j} \circ f_i$) we see that $\mathcal{T} \subseteq \mathcal{T}'$. So $\mathcal{T}$ is the initial topology w.r.t. the $f_i$.

Two useful applications, to make all this less abstract:

Subspaces of subspaces:

Let $A$ be a subspace of $B$ and $B$ a subspace of $X$ ($A \subseteq B \subseteq X$) then $A$ is a subspace of $X$ (i.e. it has the subspace topology w.r.t. $X$).

Proof: Apply the above to $i_B: B \rightarrow X$, $i_{A,B}: A \rightarrow B$, $i_A: A \rightarrow X$, all basically the identity with different domains and codomains. So by assumption $A$ has the initial topology w.r.t. $i_{A,B}$ and $B$ has the initial topology w.r.t. $i_B$. Note that $i_A = i_B o i_{A,B}$, so by the transitivity theorem (right to left) we see that $A$ has the initial topology w.r.t. $i_A$, or $A$ has the subspace topology w.r.t. $X$.

### Products and subspaces:

Let $X_i$ ($i \in I$) be a family of spaces, with subspaces $A_i \subseteq X_i$. Then $A = \prod_{i \in I} A_i$ (in the product topology of the subspace topologies) is a subspace of $X = \prod_{i \in I} X_i$ (it has the initial topology w.r.t. the inclusion).

Proof: Let $k_i$ be the inclusion mapping from $A_i$ to $X_i$. Let $k: \prod_i A_i \rightarrow \prod_i X_i$ the product mapping (as above). Note that $k$ is also the inclusion from $A$ into $X$.

Again let $p_i$ be the projections from $A$ onto the $A_i$, and $q_i$ the projections from $X$ onto the $X_i$. Then $$(\ast) q_i \circ k = k_i \circ p_i \text{ for all } i \in I \text{.}$$

$A_i$ has the initial topology w.r.t. $k_i$, and $A$ has the initial topology w.r.t. the $p_i$.

So $A$ has the initial topology w.r.t. the maps $k_i \circ p_i$ ($i \in I$) by the right to left implication of the transitivity theorem. So $A$ has the initial topology w.r.t. the maps $q_i \circ k$ by $(\ast)$. But by the transitivity theorem (the other implication ) we see that $A$ has the initial topology w.r.t. the map $k$, which is, as said, the inclusion $A \rightarrow X$. So $A$ has the subspace topology.

So we see that all these general considerations give a nice proof of "a product of subspaces is a subspace), due to the special nature of these topologies as initial topologies with respect to certain maps.

Remark: for those who know inverse limits, which are subspaces of products of a certain kind, the above theorem also shows that in fact the inverse limit topology is itself an initial topology w.r.t. the restricted projection maps. This gives rise to a canonical subbase for the inverse limit, from the existence theorem, which is sometimes useful as well.

As a final remark: a similar theory can be developped for so-called "final" topologies. This applies to situations where we have maps $f_i: X_i\rightarrow X$ where we want $X$ to have the largest topology that makes all $f_i$ continuous. Special cases include quotient topologies, disjoint sums and weak topologies induced by subspaces. Also here we have an existence theorem, a universal continuity theorem and a transitive law. See here for the details.

• Thank you for this very long and informative answer. I will take some time to get into it and get back to this question after that. – Zzz Nov 27 '15 at 16:38
• Isn't the last paragraph of your proof of the transitive law superfluous? You show the topology on $X$ (the initial topology w.r.t. the $f_i$) makes all $g_{ij}f_i$ continuous and that any topology $T'$ making all $g_{ij}f_i$ continuous is finer than the topology on $X$. And that finishes the proof. – Stefan Hamcke Nov 28 '15 at 1:20
• @StefanHamcke I see, appealing to the unicity, essentially? The last part (the forward implication) is the easier part anyway, so it doesn't really cut a lot of length. – Henno Brandsma Nov 28 '15 at 8:14
• If I'm not wrong your "Application: product maps" is similar to Theorem 2.5 in Dugundji1966. For one direction they are identical: If each $f_i$ is continuous, so also is $\prod f_i$. He has an additional condition for the reverse direction: If each $f_i$ is an open map, and all but at most finitely many are surjective, then $\prod f_i$ is also an open map. Is the additional condition "all but at most finitely many are surjective" needed only to have openness or it is also required for continuity (your case)? – PeptideChain Dec 26 '18 at 11:09
• @PeptideChain for openness only. – Henno Brandsma Dec 26 '18 at 11:15