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2) f is closed. (3) f is a homeomorphism onto a closed subspace. 8) Proposition. Let f : X → Y be continuous. (1) If f is proper, then for each subset B ⊂ Y the restriction f = fB : f −1 (B) → B is proper. (2) Let (Uj | j ∈ J) be a covering of Y such that the canonical map p : j∈J Uj → Y is a quotient map. If each restriction fj : f −1 (Uj ) → Uj is proper, then f is proper. 9) Proposition. Let f be a continuous map of a Hausdorff space X into a locally compact Hausdorff space Y . Then f is proper if and only if each compact set K ⊂ Y has a compact pre-image.

In the case that D(x) = {d ∈ D | x ∈ Ld } is empty, the infimum is, by definition, equal to 1; thus f assume the value 1 on the complement of the Ld . Proof. The sets of the form [0, a[ and ]a, 1], 0 < a < 1 form a subbasis for the topology of [0, 1]. Thus it suffices to show that their pre-images under f are open. This follows from the set-theoretic relations f −1 [0, a[ = {x|f (x) < a} = f −1 ]a, 1] = {x|f (x) > a} = (Ld | d < a} (X Ld | d > a) = (X Ld | d > a). For the proof of the last equality one uses the condition Ld ⊂ Le and the denseness of D.

6) Theorem. A map f : X → Y between topological spaces is continuous in x if and only if the image filter of each filter which converges to x converges to f (x). Proof. Let f be continuous in x and let F converge to x. Let V be a neighbourhood of f (x) and U a neighbourhood of x such that f (U ) ⊂ V . Since F converges to x we have U ∈ F and hence V ∈ f (F) since f (U ) ⊂ V . Hence f (F) is finer than U(f (x)) and converges to f (x). Let F = U(x). Each neighbourhood V of f (x) belongs to f (F) if f (F) converges to f (x).