22.8. THE DERIVATIVE AND THE CARTESIAN PRODUCT 723

Let x ∈Uy. Then (x,y) ∈U and so there exists r > 0 such that

B((x,y) ,r) ∈U.

This says that if (u,v) ∈ X×Y such that ||(u,v)− (x,y)||< r, then (u,v) ∈U. Thus if

||(u,y)− (x,y)||= ||u−x||< r,

then (u,y) ∈U. This has just said that B(x,r), the ball taken in X is contained in Uy. Thisproves the lemma.

Or course one could also consider

Ux ≡ {y :(x,y) ∈U}

in the same way and conclude this set is open in Y . Also, the generalization to many factorsyields the same conclusion. In this case, for x ∈∏

ni=1 Xi, let

||x|| ≡max(||xi||Xi

: x = (x1, · · · ,xn))

Then a similar argument to the above shows this is a norm on ∏ni=1 Xi. Consider the triangle

inequality.

∥(x1, · · · ,xn)+(y1, · · · ,yn)∥= maxi

(||xi +yi||Xi

)≤max

i

(∥xi∥Xi

+∥yi∥Xi

)≤max

i

(||xi||Xi

)+max

i

(||yi||Xi

)Corollary 22.8.2 Let U ⊆∏

ni=1 Xi be an open set and let

U(x1,··· ,xi−1,xi+1,··· ,xn) ≡ {x ∈ Fri : (x1, · · · ,xi−1,x,xi+1, · · · ,xn) ∈U} .

Then U(x1,··· ,xi−1,xi+1,··· ,xn) is an open set in Fri .

Proof: Let z ∈U(x1,··· ,xi−1,xi+1,··· ,xn). Then (x1, · · · ,xi−1,z,xi+1, · · · ,xn)≡ x ∈U by def-inition. Therefore, since U is open, there exists r > 0 such that B(x,r)⊆U. It follows thatfor B(z,r)Xi

denoting the ball in Xi, it follows that B(z,r)Xi⊆U(x1,··· ,xi−1,xi+1,··· ,xn) because

to say that ∥z−w∥Xi< r is to say that

∥(x1, · · · ,xi−1,z,xi+1, · · · ,xn)− (x1, · · · ,xi−1,w,xi+1, · · · ,xn)∥< r

and so w ∈U(x1,··· ,xi−1,xi+1,··· ,xn).Next is a generalization of the partial derivative.

Definition 22.8.3 Let g : U ⊆∏ni=1 Xi→ Y , where U is an open set. Then the map

z→ g(x1, · · · ,xi−1,z,xi+1, · · · ,xn)

is a function from the open set in Xi,

{z : x =(x1, · · · ,xi−1,z,xi+1, · · · ,xn) ∈U}