TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$

Statement

Let $(X, T_{X})$ and $(Y, T_{Y})$ be topological spaces, let $f : X \to Y$ be a function, and let $S$ be a subbasis for $T_{Y}$ (Basis and subbasis for a topology, and the topology generated by a family of sets). The following five conditions are equivalent.

(a) $f$ is continuous at every point of $X$ (Continuity of a map of topological spaces at a point and globally).
(b) $f^{- 1} [V]$ is open in $X$ for every open $V \subseteq Y$ .
(c) $f^{- 1} [F]$ is closed in $X$ for every closed $F \subseteq Y$ .
(d) $f^{- 1} [S]$ is open in $X$ for every $S \in S$ .
(e) $f [\overline{A}] \subseteq \overline{f [A]}$ for every $A \subseteq X$ , closures being taken in $X$ and in $Y$ respectively (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

Condition (d) is what makes continuity checkable against a generating family rather than against every open set, and it holds for a basis as well, a basis being in particular a subbasis for the topology it generates.

Facts & Assumptions

Given: Topological spaces $(X, T_{X})$ and $(Y, T_{Y})$ , a function $f : X \to Y$ , a subbasis $S$ for $T_{Y}$ , subsets $A \subseteq X$ and $V, F \subseteq Y$ . Preimages satisfy $f^{- 1} [Y ∖ W] = X ∖ f^{- 1} [W]$ , $f^{- 1} [⋃_{i} W_{i}] = ⋃_{i} f^{- 1} [W_{i}]$ and $f^{- 1} [⋂_{i} W_{i}] = ⋂_{i} f^{- 1} [W_{i}]$ for every family, with $f^{- 1} [Y] = X$ for the empty intersection.

[A1]

$f$ is continuous at $x$ when for every open $V ∋ f (x)$ there is an open $U ∋ x$ with $f [U] \subseteq V$ (Continuity of a map of topological spaces at a point and globally).

[A2]

A set is closed exactly when its complement is open; a set is open exactly when it is a union of open sets containing each of its points (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L1]

The topology generated by $S$ has as a basis the family $B_{S}$ of intersections of finitely many members of $S$ , the empty intersection being $Y$ ; every open set is a union of members of $B_{S}$ (A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis, Basis and subbasis for a topology, and the topology generated by a family of sets).

[L2]

$x \in \overline{A}$ if and only if every open set containing $x$ meets $A$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set, clause (c)).

[L3]

$\overline{A}$ is the smallest closed superset of $A$ , and $A$ is closed exactly when $A = \overline{A}$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

Proof

technique · direct

1.1A1A2

(a) implies (b): let $V \subseteq Y$ be open and let $x \in f^{- 1} [V]$ , so $f (x) \in V$ ; continuity at $x$ gives an open $U ∋ x$ with $f [U] \subseteq V$ , that is $U \subseteq f^{- 1} [V]$ . As $x$ was arbitrary, $f^{- 1} [V]$ is a union of open sets, hence open.

1.2A1

(b) implies (a): let $x \in X$ and let $V$ be open with $f (x) \in V$ ; then $U : = f^{- 1} [V]$ is open, contains $x$ , and satisfies $f [U] \subseteq V$ .

1.3givenA2

(b) and (c) are equivalent: $F \subseteq Y$ is closed exactly when $Y ∖ F$ is open, and $f^{- 1} [Y ∖ F] = X ∖ f^{- 1} [F]$ , so $f^{- 1} [F]$ is closed exactly when $f^{- 1} [Y ∖ F]$ is open; as $F$ ranges over the closed sets, $Y ∖ F$ ranges over the open sets.

1.4L1

(b) implies (d): every $S \in S$ is open, $S$ being contained in the topology it generates.

1.5givenL1A2

(d) implies (b): let $V \subseteq Y$ be open; by [L1] $V$ is a union of sets of the form $S_{1} \cap \dots \cap S_{n}$ with $n \geq 0$ and $S_{i} \in S$ , and $f^{- 1}$ turns unions into unions and intersections into intersections, with $f^{- 1} [Y] = X$ for $n = 0$ ; so $f^{- 1} [V]$ is a union of finite intersections of the open sets $f^{- 1} [S_{i}]$ together with $X$ , hence open.

1.6L3

(e) implies (c): let $F \subseteq Y$ be closed and put $A : = f^{- 1} [F]$ ; then $f [A] \subseteq F$ , so $f [\overline{A}] \subseteq \overline{f [A]} \subseteq \overline{F} = F$ by (e), monotonicity of the closure and [L3]; hence $\overline{A} \subseteq f^{- 1} [F] = A$ , and with $A \subseteq \overline{A}$ this gives $A = \overline{A}$ , so $A$ is closed.

2.1step 1.1L2

(b) implies (e): let $A \subseteq X$ and $x \in \overline{A}$ , and let $V$ be open with $f (x) \in V$ ; then $f^{- 1} [V]$ is open and contains $x$ , so it meets $A$ by [L2], say at $a$ ; then $f (a) \in V \cap f [A]$ , so $V$ meets $f [A]$ . As $V$ was arbitrary, $f (x) \in \overline{f [A]}$ by [L2].

3.1step 1.1step 1.2step 1.3step 1.4step 1.5step 2.1step 1.6∎

Steps 1.1 and 1.2 make (a) and (b) equivalent; step 1.3 makes (b) and (c) equivalent; steps 1.4 and 1.5 make (b) and (d) equivalent; step 2.1 gives (b) implies (e) and step 1.6 gives (e) implies (c), which closes the cycle through (c) and (b). Hence all five conditions are equivalent.

Remarks

Only (a) is pointwise. Conditions (b) to (e) are global, and none of them has a pointwise version that is equivalent to continuity at a single point: the preimage of an open set containing $f (x)$ can fail to be open while still being a neighbourhood of $x$ , which is exactly what continuity at $x$ asserts.
The inclusion in (e) may be strict for a continuous map. For the inclusion of $(0, 1)$ into $R$ and $A = (0, 1)$ , the image of the closure is $(0, 1)$ while the closure of the image is $[0, 1]$ . Equality for all $A$ is a strictly stronger condition, equivalent to $f$ being a closed map, and closed maps are defined three items below. Note that no map into a discrete space can witness strictness: there every subset is closed, so $f [\overline{A}] = f [A] = \overline{f [A]}$ always.
What the theorem does not say. It says nothing about images of open sets: a continuous map need not carry open sets to open sets, and the failure is exactly what separates a continuous bijection from a homeomorphism. That separation is recorded on this page as a false statement with an explicit two-point witness.

For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and f(A‾)⊆f(A)‾f(A)⊆f(A)​

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$