1.7+) (-, f(x)+p) x+p. 32 LA - an (3,A(=)) (5,2)-P) (1,0) ام-کم (0,0) Figure 2.6 Figure 2.5 D., Ds, and D, of Examples 2
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1.7+) (-, f(x)+p) x+p. 32 LA - an (3,A(=)) (5,2)-P) (1,0) ام-کم (0,0) Figure 2.6 Figure 2.5 D., Ds, and D, of Examples 2
2.3 Open Soto 21 3. Let (x, y) be any point of Ra, and p be any positive number. What is the relation of the D3-p-neighborhood of (x,y) to the D-p-neighborhood of (7, y)? to the D1-p-neighborhood of (x, y)? Prove that every D- and Di- neighborhood of (x, y) contains a D3-neighborhood of (x, y) and that each D3-neighborhood of (x, y) contains both a D- and a Dı-neighborhood. 4. Let X, D be the metric space as described in Example 5. Sketch the graphs of the elements of X defined by each of the following, and sketch the p-collars (Example 6) for each of these elements. a) f(t) = 1, for all 2 € (0, 1). b) f(x) = x, for all 2 € (0, 1). c) f(x) = 73, for all z € (0, 1). d) f(x) (1, if 0 < r < 5; 12, if : << 1.
Example 5. This example is given to illustrate that a metric can be defined on a set which is neither R" for some n nor a subset of R". Let X be the set of all functions from the closed interval (0, 1) into itself. If f and g are any such functions, define D(, 9) = - least upper bound {\f(x) - g(x)|| 2 € (0, 11). Since any subset of the real numbers which has an upper bound has a least upper bound and 0 < \f() - g(x) < 1 for all f,gEX and IE (0, 1), then D(5,9) is defined for all f, g EX. We will now show that D is a metric for X by showing that D satisfies each of the properties required for a metric in Definition 1: i) DC, 9) 2 0 for all f, 9 € X. Since each element of {\f(x) — 9(2)||* € [0, 1]} greater than or equal to 0, the least upper bound of this set, D(5,9), greater than or equal to 0. ii) DC, 9) = D(9,8) for all f, ge X. This follows at once from the fact that \f(x) - 9(2)) = 19() - f(x). ii) D, 9) = 0 if and only if f= g. If f= g, then f(x) = g(x) for all of 10, 11, and hence (IS(2) - (2)||& € (0,11} = {0}. Therefore it follows that D, 9) = 0. On the other hand, if Df, 9) = 0, then lub {]f(x) - g(x)||* = [0, 1]} = 0. is is