Complete the proof by filling in the blanks Proposition 4.3.14. Let (S, d) be a metric space. Then for any a and e> 0, B(x, e) is open. Proof. To show that B(x, €) is open, we must show that, for anyy e B(x, e) there exists a ô > 0 (we have to use d here since e is already defined) such that B(y, 5) C1. If this is true, then y is in the 2 of B(x, E), which would show B(x, €) C Int (B(x, E)), since y was chosen arbitrarily. Notice that d(r, y) 0. Let 5 =€ -d(x, y). Now we will show that this choice of & gives us B(y, 5) C B(x, € ). To show B(y, 8) C B(x, E), suppose ze B(y, 8). By the triangle inequality of metrics, we have d(x, z) < d(x, y) + 3< d(x,y) + 6 = d(x, y)+ € -d(x, y) = [4

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Complete the proof by filling in the blanks
Proposition 4.3.14. Let (S, d) be a metric space. Then for any a and e> 0, B(x, e) is open.
Proof. To show that B(x, e) is open, we must show that, for anyy e B(x, e) there exists a ô > 0 (we
have to use & here since e is already defined) such that B(y, 8) 1. If this is true, then y is in the 2
of B(x, e), which would show B(x, e) C Int (B(x, E)), since y was chosen arbitrarily. Notice
that d(x, y) <E. This is true by definition of B(x, e) and the fact that y e B(x, e). Hence e
-d(x, y) > 0. Let 8 =e -d(x, y). Now we will show that this choice of & gives us B(y, 8) C B(x, E
). To show B(y, 8) C B(x, e), suppose z e B(y, 8). By the triangle inequality of metrics, we have
d(x, z) < d(x, y) + 3 < d(x, y) + 6 = d(x, y)+ € -d(x, y)
4
%3!
The latter expression shows d(x, z) <E, hence z e 5. This implies 6 C B(x, €), thus B(x, E
) is 7
Transcribed Image Text:Complete the proof by filling in the blanks Proposition 4.3.14. Let (S, d) be a metric space. Then for any a and e> 0, B(x, e) is open. Proof. To show that B(x, e) is open, we must show that, for anyy e B(x, e) there exists a ô > 0 (we have to use & here since e is already defined) such that B(y, 8) 1. If this is true, then y is in the 2 of B(x, e), which would show B(x, e) C Int (B(x, E)), since y was chosen arbitrarily. Notice that d(x, y) <E. This is true by definition of B(x, e) and the fact that y e B(x, e). Hence e -d(x, y) > 0. Let 8 =e -d(x, y). Now we will show that this choice of & gives us B(y, 8) C B(x, E ). To show B(y, 8) C B(x, e), suppose z e B(y, 8). By the triangle inequality of metrics, we have d(x, z) < d(x, y) + 3 < d(x, y) + 6 = d(x, y)+ € -d(x, y) 4 %3! The latter expression shows d(x, z) <E, hence z e 5. This implies 6 C B(x, €), thus B(x, E ) is 7
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