We consider the set X = {1, 1/2, 1/3, 1/4, ...} ∪ {0} as a metric space under the Euclidean metric.
(i) Prove: for every n ∈ Natural Numbers, the singleton set {1/n} is open in X; the set {0} is not open in X.
(ii) Let Y ⊂ X. Prove: If 0 ∉ Y, then Y is open in X; if 0 ∈ Y, then Y is closed in X.
(iii) Let f: X → Reals. Prove: f is continuous iff lim f(1/n) = f(0) as n tends to infinity.
(i) Prove: for every n ∈ Natural Numbers, the singleton set {1/n} is open in X; the set {0} is not open in X.
(ii) Let Y ⊂ X. Prove: If 0 ∉ Y, then Y is open in X; if 0 ∈ Y, then Y is closed in X.
(iii) Let f: X → Reals. Prove: f is continuous iff lim f(1/n) = f(0) as n tends to infinity.
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(i) If ε < 1/(n(n+1)) then (1/n-ε, 1/n+ε) is an open set containing only 1/n, but not 1/(n-1) nor 1/(n+1).
For any ε > 0, the neighborhood (0, ε) contains all 1/n with n > 1/ε. So there are no open neighborhoods about 0 and therefore 0 must be closed. 0 is a limit point.
(ii) From (i) 0 is the only limit point, so X-{0} is open, as is any subset Y of it. If 0 ∈ Y then Y is closed under the Euclidean metric provided Y contains infinitely many members of X; and either way is closed under the subspace topology of X as 0 is the only limit point in X.
(iii) f continuous means that as x approaches a limit point p then f(x) approaches f(p). The only limit point is 0 so f(1/n) must tend to f(0) to insure continuity and conversely.
For any ε > 0, the neighborhood (0, ε) contains all 1/n with n > 1/ε. So there are no open neighborhoods about 0 and therefore 0 must be closed. 0 is a limit point.
(ii) From (i) 0 is the only limit point, so X-{0} is open, as is any subset Y of it. If 0 ∈ Y then Y is closed under the Euclidean metric provided Y contains infinitely many members of X; and either way is closed under the subspace topology of X as 0 is the only limit point in X.
(iii) f continuous means that as x approaches a limit point p then f(x) approaches f(p). The only limit point is 0 so f(1/n) must tend to f(0) to insure continuity and conversely.