6.1 CONCLUSIONS INVOLVING V FOLLOWED BY 3 195S n T. The key is the choice of 6; we can go no further in the proof
until we've decided what 6 should be.
Now what do our hypotheses tell us? Since S is open and x E S, then
x is an interior point of S. (Note the implicit use of "specialization."
Every point of S is an interior point of S; hence our particular point x
must be an interior point.) Similarly, x is an interior point of T. Using
the definition of "interior point" twice, we see that there exist positive
real numbers 6, and 6, such that N(x; 6,) r S and N(x; 6,) E T. Ex-
amples 7 and 8 may have conditioned you to expect that we will define
the desired 6 in terms of 6, and 6,. (Perhaps we should say that if you
have already made a mental note that we will doubtless define 6 in terms
of 6, and a,, then you're making good progress.) But the question is,
"How to do it?'Let us draw some pictures, such as those shown in
Figure 6.1.
Figure 6.lb illustrates the two values of 8, and 6,. Any point within
6, of x is inside S; any point within 6, of x is inside T. We want to
choose 6 small enough (but still positive) so that any point within 6 of x
will be inside both S and T. Finally, then, how will we choose S? Why
not let 6 be the smaller of 6, and d2? In symbols 6 = min (6 ,, d2)! This
choice seems reasonable; if it is correct, we ought to be able to complete
the proof by showing N(x; 6) c S n T. To do this, in turn, we recall
the choose method and let y G N(x; 6). We claim that y E S n T. Now
Ix - yl < 6 s 6, so that y E N(x; 6,) c S so y E S. Also, lx - yl < 6 s 6,
so that y E N(x; 6,) c T. Hence y E S n T so that N(x; 6) c S n T, and
our proof is complete.Figure 6.1 A picture suggests how to complete the proof that
the intersection of two open sets is open. (a) A picture of a
two-dimensional version of the situation described in Example 9.
(6) In this case, we take 6 = a,, since 6, < 6,.