[:finite product topology:] [:product topology:] • A product topology on XX and YY (both with an implciit topology) is the topology X×YX \times Y generated by the topological basis B\mathscr B given by sets of the form U×VU \times V where UU is open in XX and VV is open in YY • Intuition: consider X=Y=RstdX = Y = \mathbb R_\text{std}. Elements of B\mathscr B are then open rectangles with closed horizontal and vertical stripes taken out (wish I could draw on Roam lol) • Note: This definition generalizes to to finite products of topologies, but not to infinite products: then there are multiple canonical definitions (infinite product topology, box topology) • Note that B\mathscr B is not itself a topology. Let X=Y=RstdX = Y = \mathbb R_\text{std}. Then (02)2(0\dots 2)^2 and (13)2(1\dots 3)^2 are both in B\mathscr B, and so is their intersection, but their union is not • Here I use (ab)(a \dots b) to mean the interval (a,b)(a, b), to avoid ambiguous notation • Proof that B\mathscr B is a basis: • Condition 1: B\mathscr B covers X×YX \times Y • Intuition: true since TX\mathscr T_X covers XX and TY\mathscr T_Y covers YY • Take an x,yX×Y\langle x, y\rangle \in X \times Y. WTS that x,y\langle x, y\rangle is in some basis element • Note that there is some UxU_x open in XX containing xx, and some UyU_y open in YY containing yy • Note that Ux×UyU_x \times U_y contains x,y\langle x, y\rangle and is a basis element. Done. • Condition 2: given B1,B2BB_1, B_2 \in \mathscr B and xB1B2x \in B_1 \cap B_2, exists some BBB' \in \mathscr B s.t. BB1B2B' \subseteq B_1 \cap B_2 and xBx \in B' • Take B1,B2BB_1, B_2 \in \mathscr B and xB1B2x \in B_1 \cap B_2 • Let B=B1B2B' = B_1 \cap B_2 • Then clearly xBx \in B' • What remains to show is that BB' is a basis element. • Write basis elements B1B_1 as U1×V1U_1 \times V_1 and B2B_2 as U2×V2U_2 \times V_2 • Then B=(U1×V1)(U2×V2)=(U1U2)×(V1V2)B' = (U_1 \times V_1) \cap (U_2 \times V_2) = (U_1 \cap U_2) \times (V_1 \cap V_2)U1U2U_1 \cap U_2 and V1V2V_1 \cap V_2 are both open in their respective spaces, so BB' is a basis element • • Note that if B\mathscr B is a basis for TX\mathscr T_X and C\mathscr C is a basis for TY\mathscr T_Y then the collection D={B×C:BB,CC}\mathscr D = \{ B \times C : B \in \mathscr B, C \in \mathscr C \} is a basis for X×YX \times Y • Proof: elided (for now?) • • The functions π1=(x,y)x\pi_1 = (x, y) \mapsto x and π2=(x,y)y\pi_2 = (x, y) \mapsto y are called projections of X×YX \times Y • Then π11(x0)={x,yX×Y:x=x0}\pi_1^{-1}(x_0) = \{ \langle x, y \rangle \in X \times Y : x = x_0 \} • Read: π11(x)\pi_1^{-1}(x) are all those 2-tuples where the first element is xx • Intuition: for X=Y=RX = Y = \mathbb R, π11(1)\pi_1^{-1}(1) is the vertical line x=1x = 1 • And for the interval I=(x0,xf)I = (x_0, x_f), π11(I)\pi_1^{-1}(I) is the vertical strip from x=x0x = x_0 through x=xfx = x_f • Read: π11(U)\pi_1^{-1}(U) are all those 2-tuples where the first elements is in UU • ![](https://firebasestorage.googleapis.com/v0/b/firescript-577a2.appspot.com/o/imgs%2Fapp%2Fplace%2Fiu6t-mSzvo.png?alt=media&token=48969a03-22ca-43ba-af8c-3e7fbe388600) • We can define a topological subbasis S\mathcal S for X×YX \times Y as follows: • Let Vert={π11(V):V open in Y}\text{Vert} = \{ \pi_1^{-1}(V) : V \text{ open in } Y \} • Let Horiz={π21(U):U open in X}\text{Horiz} = \{ \pi_2^{-1}(U) : U \text{ open in } X \}S=HorizVert\mathcal S = \text{Horiz} \cup \text{Vert} • I will show that S\mathcal S is a subbasis for X×YX \times Y by showing that elements of the previously-discussed B\mathcal B are intersections of S\mathcal S. Also see subbasis to basis. • Take some BBB \in \mathcal B • Then B=U×VB = U \times V for some UU open in XX and VV open in YY • Claim: B=π11(U)π21(V)B = \pi_1^{-1}(U) \cap \pi_2^{-1}(V) • Will prove by showing x,yB    x,yπ11(U)π21(V)\langle x, y \rangle \in B \iff \langle x, y \rangle \in \pi_1^{-1}(U) \cap \pi_2^{-1}(V)()(\Longrightarrow) • Take x,yB\langle x, y \rangle \in B • Then xUx \in U and yVy \in V • Since xUx \in U then x,yπ11(U)\langle x, y \rangle \in \pi_1^{-1}(U) • Since yVy \in V then x,yπ21(V)\langle x, y \rangle \in \pi_2^{-1}(V) • Thus x,yπ11(U)π21(V)\langle x, y \rangle \in \pi_1^{-1}(U) \cap \pi_2^{-1}(V)()(\Longleftarrow) • Take x,yπ11(U)π21(V)\langle x, y \rangle \in \pi_1^{-1}(U) \cap \pi_2^{-1}(V) • Then xUx \in U and yVy \in V • So x,yU×V\langle x, y \rangle \in U \times V Referenced by: