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Theorem 18.1. Let X and Y be topological spaces; let $f: X \to Y$. Then the following are equivalent: (1) $f$ is continuous. (2) For every subset A of X, one has $f(\bar{A}) \subset \overline{f(A)}$. (3) For every closed set B of Y, the set $f^{-1}(B)$ is closed in X. (4) For each $x \in X$ and each neighborhood V of $f(x)$, there is a neighborhood U of x such that $f(U) \subset V$. If the condition in (4) holds for the point x of X, we say that $f$ is continuous at the point x. Proof. We show that $(1) \implies (2) \implies (3) \implies (1)$ and that $(1) \implies (4) \implies (1)$. $(1) \implies (2)$. Assume that $f$ is continuous. Let A be a subset of X. We show that if $x \in \bar{A}$, then $f(x) \in \overline{f(A)}$. Let V be a neighborhood of $f(x)$. Then $f^{-1}(V)$ is an open set of X containing x; it must intersect A in some point y. Then V intersects $f(A)$ in the point $f(y)$, so that $f(x) \in \overline{f(A)}$, as desired. $(2) \implies (3)$. Let B be closed in Y and let $A = f^{-1}(B)$. We wish to prove that A is closed in X; we show that $\bar{A} = A$. By elementary set theory, we have $f(A) =$ $f(f^{-1}(B)) \subset B$. Therefore, if $x \in \bar{A}$, $f(x) \in f(\bar{A}) \subset \overline{f(A)} \subset \bar{B} = B$, so that $x \in f^{-1}(B) = A$. Thus $\bar{A} \subset A$, so that $\bar{A} = A$, as desired. $(3) \implies (1)$. Let V be an open set of Y. Set $B = Y - V$. Then $f^{-1}(B) = f^{-1}(Y) - f^{-1}(V) = X - f^{-1}(V)$. Now B is a closed set of Y. Then $f^{-1}(B)$ is closed in X by hypothesis, so that $f^{-1}(V)$ is open in X, as desired.

          Theorem 18.1. Let X and Y be topological spaces; let $f: X \to Y$. Then the
following are equivalent:
(1) $f$ is continuous.
(2) For every subset A of X, one has $f(\bar{A}) \subset \overline{f(A)}$.
(3) For every closed set B of Y, the set $f^{-1}(B)$ is closed in X.
(4) For each $x \in X$ and each neighborhood V of $f(x)$, there is a neighborhood U
of x such that $f(U) \subset V$.
If the condition in (4) holds for the point x of X, we say that $f$ is continuous at
the point x.
Proof. We show that $(1) \implies (2) \implies (3) \implies (1)$ and that $(1) \implies (4) \implies (1)$.
$(1) \implies (2)$. Assume that $f$ is continuous. Let A be a subset of X. We show that if
$x \in \bar{A}$, then $f(x) \in \overline{f(A)}$. Let V be a neighborhood of $f(x)$. Then $f^{-1}(V)$ is an open
set of X containing x; it must intersect A in some point y. Then V intersects $f(A)$ in
the point $f(y)$, so that $f(x) \in \overline{f(A)}$, as desired.
$(2) \implies (3)$. Let B be closed in Y and let $A = f^{-1}(B)$. We wish to prove that A
is closed in X; we show that $\bar{A} = A$. By elementary set theory, we have $f(A) =$
$f(f^{-1}(B)) \subset B$. Therefore, if $x \in \bar{A}$,
$f(x) \in f(\bar{A}) \subset \overline{f(A)} \subset \bar{B} = B$,
so that $x \in f^{-1}(B) = A$. Thus $\bar{A} \subset A$, so that $\bar{A} = A$, as desired.
$(3) \implies (1)$. Let V be an open set of Y. Set $B = Y - V$. Then
$f^{-1}(B) = f^{-1}(Y) - f^{-1}(V) = X - f^{-1}(V)$.
Now B is a closed set of Y. Then $f^{-1}(B)$ is closed in X by hypothesis, so that $f^{-1}(V)$
is open in X, as desired.

Theorem 18.1. Let X and Y be topological spaces; let f: X → Y. Then the
following are equivalent:
(1) f is continuous.
(2) For every subset A of X, one has f(A̅) ⊂f(A).
(3) For every closed set B of Y, the set f^-1(B) is closed in X.
(4) For each x ∈ X and each neighborhood V of f(x), there is a neighborhood U
of x such that f(U) ⊂ V.
If the condition in (4) holds for the point x of X, we say that f is continuous at
the point x.
Proof. We show that (1) (2) (3) (1) and that (1) (4) (1).
(1) (2). Assume that f is continuous. Let A be a subset of X. We show that if
x ∈A̅, then f(x) ∈f(A). Let V be a neighborhood of f(x). Then f^-1(V) is an open
set of X containing x; it must intersect A in some point y. Then V intersects f(A) in
the point f(y), so that f(x) ∈f(A), as desired.
(2) (3). Let B be closed in Y and let A = f^-1(B). We wish to prove that A
is closed in X; we show that A̅ = A. By elementary set theory, we have f(A) =
f(f^-1(B)) ⊂ B. Therefore, if x ∈A̅,
f(x) ∈ f(A̅) ⊂f(A)⊂B̅ = B,
so that x ∈ f^-1(B) = A. Thus A̅⊂ A, so that A̅ = A, as desired.
(3) (1). Let V be an open set of Y. Set B = Y - V. Then
f^-1(B) = f^-1(Y) - f^-1(V) = X - f^-1(V).
Now B is a closed set of Y. Then f^-1(B) is closed in X by hypothesis, so that f^-1(V)
is open in X, as desired.

Added by Virginia B.

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