Math 140A Homework 10 (Solutions)
نویسنده
چکیده
Solution. Suppose p ∈ E. If p ∈ E, then f(p) ∈ f(E) ⊆ f(E). Otherwise, p is a limit point of E, and we claim that if f(p) / ∈ f(E), then f(p) is a limit point of f(E). Let ε > 0 be given, and choose δ > 0 such that for all q ∈ X, if 0 < dX(p, q) < δ, then dY (f(p), f(q)) < ε. Since p is a limit point of E, there exists a q ∈ E such that 0 < dX(p, q) < δ. Then f(q) ∈ f(E) (so in particular f(q) 6= f(p)) and dY (f(p), f(q)) < ε. Thus, f(p) is a limit point of f(E), so again f(p) ∈ f(E). We conclude that f(E) ⊆ f(E). To see that we might have f(E) 6= f(E), consider X = R with the discrete metric and Y = R with the usual metric. Let E = Q, and let f : X → Y be the identity map: f(x) = x for all x ∈ X. Note that f is continuous (since every subset of X is open, the preimage of every subset of Y is open in X). Then E = Q in X, so f(E) = Q in Y , but f(E) = Q = R in Y .
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تاریخ انتشار 2017