Let (X,d) be a metric space where a is an element of X and r > 0. Show that:
the closure of {x: d(x,a) < r} is a proper subset of {x: d(x,a) <= r}.
Give an example in which the sets differ.
the closure of {x: d(x,a) < r} is a proper subset of {x: d(x,a) <= r}.
Give an example in which the sets differ.
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The question must be: "is a subset of" and not "is a proper subset of". Because there are metric spaces in which the two sets are the same (e.g. the real line with the usual metric d(x,y) = |x - y|). And if the sets were always different, it probably wouldn't ask you to give an example where they are different.
Fix any a and r and any y in the closure of {x in X: d(x,a) < r}. There is necessarily a sequence of points (x_n) in {x in X: d(x,a) < r} that converges to y (this is either the definition of "closure", or one of the elementary consequences of it). So we have that (x_n) converges to y, and that d(x_n, a) < r for all n.
Fix any e > 0. As (x_n) converges to y, there is a positive integer N with the property that whenever n >= N one has d(x_n, y) < e. In particular we have d(x_N, y) < e, and hence
d(a,y) <= d(a,x_N) + d(x_N,y) < r + e
(by the triangle inequality, the property of the sequence (x_n), and the choice of N). Conclusion: for all e > 0 we have d(a,y) < r + e.
It follows that d(a,y) <= r (otherwise we could get a contradiction by taking e to be the positive number d(a,y) - r in the above argument). Since y was an arbitrary element of {x in X: d(x,a) < r} we conclude that {x in X: d(x,a) < r} is a subset of {x in X: d(x,a) <= r}.
For an example where the sets differ, consider any set X with more than two elements, and let d be the "discrete" metric on X, that is, define
d(x,y) = 0 if x = y and 1 otherwise.
Fixing any a in X you can see that {x in X: d(x,a) < 1} = {x in X: d(x,a) = 0} = {x in X: x = a} = {a}, whereas {x in X: d(x,a) <= 1} is all of X. Since X has more than two elements, X and {a} are not the same.
Fix any a and r and any y in the closure of {x in X: d(x,a) < r}. There is necessarily a sequence of points (x_n) in {x in X: d(x,a) < r} that converges to y (this is either the definition of "closure", or one of the elementary consequences of it). So we have that (x_n) converges to y, and that d(x_n, a) < r for all n.
Fix any e > 0. As (x_n) converges to y, there is a positive integer N with the property that whenever n >= N one has d(x_n, y) < e. In particular we have d(x_N, y) < e, and hence
d(a,y) <= d(a,x_N) + d(x_N,y) < r + e
(by the triangle inequality, the property of the sequence (x_n), and the choice of N). Conclusion: for all e > 0 we have d(a,y) < r + e.
It follows that d(a,y) <= r (otherwise we could get a contradiction by taking e to be the positive number d(a,y) - r in the above argument). Since y was an arbitrary element of {x in X: d(x,a) < r} we conclude that {x in X: d(x,a) < r} is a subset of {x in X: d(x,a) <= r}.
For an example where the sets differ, consider any set X with more than two elements, and let d be the "discrete" metric on X, that is, define
d(x,y) = 0 if x = y and 1 otherwise.
Fixing any a in X you can see that {x in X: d(x,a) < 1} = {x in X: d(x,a) = 0} = {x in X: x = a} = {a}, whereas {x in X: d(x,a) <= 1} is all of X. Since X has more than two elements, X and {a} are not the same.