Finite intersection of open sets is open?

Let x be in S. Then x is in S_k for all i. Since each S_k is open, there exist r_k such that for each k, the open interval (x - r_k, x + r_k) is contained in S_k. Let r = min{r_1,...,r_n}. For all y in R, if |x - y| < r, then |x - y| < r_k for each k, implying y is in S_k for all k, i.e., y is in S. Thus, the open interval (x - r, x + r) is contained in S. Since x was arbitrary, S is open.

Well, see link for definition of an open set : (i hope you know it !! ;-) ) http://en.wikipedia.org/wiki/Open_set then : for n = 2 : the intersection of two open sets in an open set : easy to prove : let be d(x,y) the metric ------> can be : d(x , y) = | x - y | for any x € S1 ∩ S2 there exists ε1 so that for any y, if d(x,y) < ε1 then y € S1 and the same way, there exists ε2 so that for any y, if d(x,y) < ε2 then y € S2 therefore, if we define : ε = min ( ε1 , ε2 ) then : if d(x,y) < ε then y € S1 and y € S2 therefore : y € S1 ∩ S2 conclusion : S1 ∩ S2 is a closed set induction hyp : if for any n , S1, S2, ..., Sn being closed set, we have : ∩ ( i from 1 to n of ) Si (is a short notation for S1 ∩ S2 ... ∩ Sn) is a closed set : then , if S_(n+1) is a closed set , [ ∩ ( i from 1 to n of ) Si ] is a closed set and S_(n+1) is a closed set therefore , according to the induction hyp. , [ ∩ ( i from 1 to n of ) Si ] ∩ S_(n+1) = S1 ∩ S2 ∩ ... ∩ Sn ∩ S_(n+1) is a closed set finally : the intersection of a finite collection of open sets is an open set. NB: wihch is NOT the case for an infinite collection : the intersection of an infinite collection of open sets can be a CLOSED set : example : ------------> [ ∩ ( i from 1 to oo of ) ( -1/n , 1/n) ] = { 0 } is closed !! ;-) hope it' ll help !! PS: pls don't forget to give Best Answers, because according to "new rules" :( only the asker may give Best Answers. To me or to anybody else !! ---> it's about keeping people motivated to answer! michael