Nonexamples:: *[[Nonexamples]]* Constructions:: *[[Constructions|Used in the construction of...]]* Specializations:: *[[Specializations]]* Generalizations:: *[[Generalizations]]* Justifications and Intuition:: *[[Justifications and Intuition]]* Examples:: *[[standard topology on the real plane]]* Properties:: [[product of factor bases is basis of product topology]] Sufficiencies:: *[[Sufficiencies]]* Equivalences:: *[[Equivalences]]* ---- > [!definition] Definition. ([[product topology]]) > Let $X$ and $Y$ be [[topological space|topological spaces]]. The **product topology** on $X \times Y$ is the [[topological space|topology]] having as [[basis for a topology|basis]] the collection $\mathscr{B}$ of all sets of the form $U \times V$, where $U$ is an [[open set|open subset]] of $X$ and $V$ is an [[open set|open subset]] of $Y$. - [ ] def. with universal property - [ ] relate the above to the 'coarsest topology s.t. coordinate projections are continuous' ([[initial topology]]) > [!definition] Definition. ([[product topology]]) > More generally, let $\{ X_{\beta} \}_{\beta \in J}$ be an indexed family of [[topological space|topological spaces]]. Let $\mathscr{S}_{\beta}$ denote the collection $\mathscr{S}_{\beta}:= \{ \pi_{\beta}^{-1} (U_{\beta}): U_{\beta} \text{ open in } X_{\beta} \} \subset \prod_{\beta}^{}X_{\beta} ,$ where $\pi_{\beta}: \prod_{\beta}^{}X_{\beta} \to X_{\beta}$ is the [[projection function|projection mapping]] associated with the index $\beta$. Define $\mathscr{S} := \bigcup_{\beta \in J}^{} \mathscr{S}_{\beta}.$ The [[subbasis for a topology|topology generated by the subbasis]] $\mathscr{S}$ is called the **product topology** on the **product space** $\prod_{\alpha}^{}X_{\alpha}$. \ Equivalently (see the justification below), we can characterize the **product topology** as the [[topology generated by a basis|topology generated by the basis]] $\mathscr{B}= \left\{ \textcolor{Skyblue}{\prod_{\alpha \in I}^{} U_{\alpha}} \text{ s.t. } U_{\alpha} \text{ open in } X_{\alpha} \text{ and } \{\alpha \in J : U_{\alpha} \neq X_{\alpha}\} \text{ is finite} \right\}.$ Elements of this [[basis for a topology|basis]] are called **cylinder sets**. > [!note] Remark. > It is clear that the [[box topology]] and [[product topology]] agree for finite products. For a number of reasons, the [[product topology]] is preferred in infinite dimensions. So, whenever one sees the notation $\prod_{\alpha}^{}X_{\alpha}$ they should assume it refers to the product endowed with the [[product topology]]. > [!justification] > Let's consider the [[basis for a topology|basis]] $\mathscr{B}$ that $\mathscr{S}$ [[subbasis for a topology|generates]]. $\mathscr{B}$ is comprised of all finite intersections of elements of $\mathscr{S}$. If we intersect elements belonging to the same one of the sets $\mathscr{S}_{\beta}$, we do not get anything new, since $\pi_{\beta}^{-1} (U_{\beta}) \cap \pi_{\beta}^{-1}(V_{\beta}) = \pi_{\beta}^{-1}(U_{\beta} \cap V_{\beta}).$ We get something new only when we intersect elements from different sets $\mathscr{S}_{\beta}$. The typical element $B$ of the basis $\mathscr{B}$ can thus be described as follows: Let $\beta_{1},\dots, \beta_{n}$ be a finite set of distinct indices from the index set $J$, and let $U_{\beta_{i}}$ be an open set in $X_{\beta_{i}}$, for $i=1,\dots,n$. Then $B= \bigcap_{i=1}^{n} \pi_{\beta_{i}}^{-1}(U_{\beta_{i}})$ is the typical element of $\mathscr{B}$. Now a point is in $B$ if and only if its $\beta_{1}^{}$th coordinate is in $U_{\beta_{1}}$, its $\beta_{2}$th coordinate in $U_{\beta_{2}}$, and so on. There is no restriction on whatever the $\alpha$th coordinate of the point is if $\alpha$ is not one of the indices $\beta_{1},\dots,\beta_{n}$. As a result, we can write $B$ as the product $B=\prod_{\alpha \in J}^{}U_{\alpha}$ where $U_{\alpha}$ denotes the entire space $X_{\alpha}$ if $\alpha \notin \{ \beta_{1},\dots, \beta_{n} \}$. > [!justification] Two-Set Justification. > We must verify that $\mathscr{B}$ is indeed a [[basis for a topology|basis]]. Covering is trivial since $X \times Y$ is itself a [[basis for a topology|basis]] element. Let $(x,y) \in X \times Y$ belong to the intersection of elements in $\mathscr{B}$: $(x,y) \in (U_{1} \times V_{1}) \cap (U_{2} \times V_{2}).$ > Then [[nestles in|nestling]] is satisfied because $(U_{1} \times V_{1}) \cap (U_{2} \times V_{2})$ is itself a [[basis for a topology|basis]] element since $(U_{1} \times V_{1}) \cap (U_{2} \times V_{2}) = (U_{1} \cap U_{2}) \times (V_{1} \cap V_{2})$ > and $U_{1} \cap U_{2}$, $V_{1} \cap V_{2}$ must be [[open set|open in]] $X,Y$ respectively. > > > [!basicexample] > Let $I:=[0,1]$. Compare the (1) [[product topology]] on $I \times I$, where $I$ inherits the [[standard topology on the real line|standard topology]] from $\mathbb{R}$, (2) the [[dictionary order relation|dictionary order]] [[order topology|topology]] on $I \times I$, and (3) the [[subspace topology|topology]] $I \times I$ [[subspace topology|inherits as a subspace]] of $\mathbb{R} \times \mathbb{R}$ in the [[dictionary order relation|dictionary order]] [[order topology|topology]]. Denote the respective [[topological space|topologies]] $\tau_{1}, \tau_{2}, \tau_{3}$. The claim is $\tau_{1} \subset \tau_{2} \subset \tau_{3}$. \ We first show that $\tau_{3}$ is *strictly* [[comparable topologies|finer than]] $\tau_{2}$. To show this, we can show that the [[basis for the subspace topology|basis]] $\mathscr{B}_{3}$ which generates $\tau_{3}$ contains the [[order topology|basis]] $\mathscr{B}_{2}$ which generates $\tau_{2}$. So let $B \in \mathscr{B}_{2}$; note that $B$ is either an [[open interval]] in $I^{2}$ or a [[half-open interval]] in $I^{2}$ containing one of the endpoints $(0,0)$ or $(1,1)$. In the first case $B$ is open in $\mathbb{R}^{2}$ and we immediately write $B=I^{2} \cap B$ implying $B$ [[basis for the subspace topology|is a member of]] $\mathscr{B}_{3}$. In this second case WLOG $B=$ $\left\{ (x,y) : (0,0) \leq (x,y) < \left( a,b \right) \right\}$ for some point $(a,b) \in I^{2}$. This is the intersection of $I^{2}$ with [[dictionary order relation|basis element]] $\left\{ (x,y) : (-1,0) \leq (x,y) < \left( a,b \right) \right\}$ of dict-ordered $\mathbb{R}^{2}$. Thus it is an element of $\mathscr{B}_{3}$, implying $\tau_{2} \subset \tau_{3}$. \ However, $\tau_{3} \not \subset \tau_{2}$. To see this, consider the [[open set]] in $\tau_{3}$ defined as $O:= \left\{ \frac{1}{2} \right\} \times [0,1], $ which is [[open set|open in]] $\tau_{3}$ because it equals $I_{2} \cap \left\{ \frac{1}{2} \right\} \times (-2,2)$. \ Next we show that $\tau_{2}$ is strictly finer than $\tau_{1}$. Clearly $\tau_{2} \not \subset \tau_{1}$— there exist sets [[open set|open in]] the [[order topology]] on $I^{2}$ but not in the [[product topology]] on $I^{2}$. One example is $\left\{ \frac{1}{2} \right\} \times (0,1)$. On the other hand, any set in $\mathscr{B}_{1}$ is a product $U \times V$ of [[interval]]s in $I$, which may be represented as an [[interval]] in the [[dictionary order relation|dictionary order]] and hence an element of $\mathscr{B}_{2}$. ^58ba6d > [!basicexample] > Let $(x_{n}) \to x$ be a [[converge|convergent]] sequence in [[topological space]] $X$, and $(y_{n}) \to y$ a [[converge|convergent]] [[sequence]] in [[topological space]] $Y$. Then $(x_{n},y_{n})_{n \in \mathbb{N}} \to (x,y)$ in $X \times Y$. To see this, by [[converge#^711ef2|this equivalence]] it suffices to show that for any [[basis for a topology|basis element]] $U \times V \subset X \times Y$, $U,V$ open subsets of $X,Y$ respectively, we can find $N \in \mathbb{N}$ s.t. for all $n > N$ we have $(x_{n},y_{n}) \in U \times V$. Since $(x_{n}) \to x$ we can obtain $N_{x}$ s.t. for all $n > N_{x}$ we have $x_{n} \in U$. We can obtain $N_{y}$ in a precisely similar fashion. Then setting $N:=\max(N_{x}, N_{y})$, it is clear that for all $n > N$ we have both $x_{n} \in U$ and $y_{n} \in V$; hence $(x_{n}, y_{n}) \in U \times V$. > [!basicexample] > Let $X=\mathbb{R}^{\mathbb{R}}$ be equipped with the [[product topology]], and be identified with the set of functions $f:\mathbb{R} \to \mathbb{R}$. Prove that the set of positive functions $A=\{ f: \mathbb{R} \to \mathbb{R} : \forall x \in \mathbb{R}, f(x) > 0 \}$ > is not [[topological space|open in]] $\mathbb{R}^{\mathbb{R}}$ with the [[product topology]]. > From [[product of factor bases is basis of product topology]], we know that the [[product topology]] is [[topology generated by a basis|generated by]] a basis consisting of the collection of sets of the of the form $\prod_{\alpha \in J}^{}I_{\alpha},$ where $I_{\alpha}$ is a bounded [[open interval]] in $\mathbb{R}$ for finitely many indices and $I_{\alpha}=(-\infty, \infty)=\mathbb{R}$ for all the remaining indices. But this means that, given a positive function $f \in \mathbb{R}^{\mathbb{R}}$, any basis element $B_{f}$ containing $f$ must feature a copy of $\mathbb{R}$ as a factor: $I_{\alpha'}=\mathbb{R}$ for some $\alpha' \in J$ . But $B_{f}$ thus contains a function that is not positive — for example, $B_{f}$ contains the function $g$ that is defined to match $f$ for all indices $\alpha \neq \alpha'$ but equal $0$ for the index $\alpha'$. ^e123eb ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```