2.7Continuous Functions

We prove $(1) \Longrightarrow (2) \Longrightarrow (3) \Longrightarrow (1)$ and $(1) \Longrightarrow (4) \Longrightarrow (1)$.

$(1) \Longrightarrow (2)$: Suppose $f$ is continuous. We seek to show that, for any $x \in \bar{A}$, it is true that $f(x) \in \bar{f(A)}$. To get started, consider an open neighborhood $V \subseteq Y$ about $f(x) \in f(\bar{A})$. Then note $f^{-1}(V) \subseteq X$ is open by continuity. Since $x \in \bar{A}$, all open neighborhoods of $x$ intersect $A$. Since $x \in f^{-1}(V)$ and $f^{-1}(V)$ is open, it follows that there is some $y \in f^{-1}(V) \cap A$. Then note $f(y) \in V \cap f(A)$. Thus arbitrary open neighborhoods of $f(x)$ intersect $f(A)$, implying $f(x)$ is in the closure of $f(A)$. Thus $f(x) \in \bar{f(A)}$, and ultimately $f(\bar{A}) \subseteq \bar{f(A)}$.

$(2) \Longrightarrow (3)$: Suppose that, for every $A \subseteq X$, it is true that $f(\bar{A}) \subseteq \bar{f(A)}$. Let $B \subseteq Y$ be closed, and let $A = f^{-1}(B)$—we wish to show $A$ is closed, and we do this by showing $A = \bar{A}$. Trivially, we note $A \subseteq \bar{A}$, so we show $\bar{A} \subseteq A$. As such, let $x \in \bar{A}$. Then note

Thus $x \in f^{-1}(B) = A$, meaning $A = \bar{A}$, meaning $f^{-1}(B)$ is closed.

$(3) \Longrightarrow (1)$: Suppose that, for every closed set $B \subseteq Y$, the preimage $f^{-1}(B) \subseteq X$ is closed as well. Let $O \subseteq Y$ be open. Then note $Y \setminus O$ is closed in $Y$, and so $f^{-1}(Y \setminus O)$ is closed in $X$. Thus $X \setminus f^{-1}(Y \setminus O) = f^{-1}(O)$ is open, meaning $f$ is continuous.

$(1) \Longrightarrow (4)$: Suppose $f$ is continuous. Fix $x \in X$, then let $V \subseteq Y$ be an open neighborhood of $f(x)$. Since $f$ is continuous, $U = f^{-1}(V) \subseteq X$ is an open neighborhood of $x$. Then note $f(U) = f(f^{-1}(V)) \subseteq V$ as desired.

$(4) \Longrightarrow (1)$: Let $V \subseteq Y$ be an arbitrary open set. We seek to show $f^{-1}(V) \subseteq X$ is open. If $f^{-1}(V) = \emptyset$, then we are done. Otherwise, let $x \in f^{-1}(V)$. Then $f(x) \in V$, but there is some open neighborhood $U_x \subseteq X$ of $x$ such that $f(U_x) \subseteq V$. Note then that $U_x \subseteq f^{-1}(V)$ for each $x \in f^{-1}(V)$, and so $\bigcup_{x \in f^{-1}(V)} U_x \subseteq f^{-1}(V)$. Moreover, $\{U_x\}_{x \in f^{-1}(V)}$ covers $f^{-1}(V)$, so $f^{-1}(V) \subseteq \bigcup_{x \in f^{-1}(V)} U_x$. Thus,

which is the arbitrary union of open sets, completing the proof. ❦

Let $O \subseteq Y$ be a closed subset of $Y$. Then note

By continuity, note $f^{-1}(O)$ is closed in $A$. By definition, this means $f^{-1}(O) = C \cap A$, where $C$ is some closed subset of $X$. But since $A$ is also a closed subset of $X$, this makes $f^{-1}(O)$ a closed subset of $X$. An entirely symmetric argument reveals $g^{-1}(O)$ is closed as well. Thus $h^{-1}(O)$ is closed in $X$, completing the proof. ❦

$(\Longrightarrow)$: Suppose $f$ is continuous. Let $U \subseteq X$ be open: we seek to show $f_X^{-1}(U)$ is open as well. Note that $f_X = \pi_X \circ f$, so then $f_X^{-1} = (\pi_X \circ f)^{-1} = f^{-1} \circ (\pi_X)^{-1}$. Note that $\pi_X^{-1}(U) = U \times Y$, which is an open subset of $X \times Y$ when gifted with the product topology. Then, by continuity of $f$, the preimage of $U \times Y$ is open, meaning $f_X^{-1}$ maps open sets to open sets, making $f_X$ continuous. An entirely symmetric argument proves $f_Y$ is continuous as desired.

$(\Longleftarrow)$: Suppose $f_X, f_Y$ are continuous. Let $U \subseteq X \times Y$ be an open: we seek to show $f^{-1}(U)$ is open. Let $U = \bigcup_{\lambda \in \Lambda} (X_\lambda \times Y_\lambda)$, where $\{X_\lambda\}_{\lambda \in \Lambda}$ is a family of open subsets of $X$ and $\{Y_\lambda\}_{\lambda \in \Lambda}$ is a family of open subsets in $Y$. Now note

By continuity of $f_X$ and $f_Y$, we note each $f_X^{-1}(X_\lambda) \cap f_Y^{-1}(Y_\lambda)$ is the intersection of two open sets, and is therefore open. Because $f^{-1}(U)$ is a union of such open sets, it is also open, meaning $f$ is continuous. ❦

Suppose $x$ is a limit point of $A$. Let $V \subseteq Y$ be some open neighborhood of $f(x)$. By continuity of $f$, note there is some open neighborhood $U \subseteq X$ of $x$ such that $f(U) \subseteq V$. Then note $U \cap (A \setminus \{x\}) \ne \emptyset$ by the definition of a limit point. If our scenario is such that $f(U \cap (A \setminus \{x\})) = \{f(x)\}$ for every single open neighborhood of $x$, then $f(x)$ is not a limit point of $f(A)$.

Let $U = X$. Then $f(U \cap (A \setminus \{x\})) = f(A \setminus \{x\}) = \{f(x)\}$ would make what we seek false. A candidate for disproof is a constant function. Let $f: \bR \to \bR$ be defined as $f(x) = 0$, then let $A = [0, 1] \subseteq \bR$. Of course, $0$ is a limit point for $A$, but $f(A) = \{0\}$, and singletons don't have limit points, completing the disproof.

Note $f(X) = X \times \{y_0\}$, so we seek to show $\tilde{f}: X \to X \times \{y_0\}$ is a homeomorphism, where the codomain is blessed with the subspace topology. $\tilde{f}$ is obviously bijective, so we simply show $\tilde{f}$ and $\tilde{f}^{-1}$ are continuous.

First, suppose $V \subseteq X \times \{y_0\}$ is open. Then $V = V' \cap (X \times \{y_0\})$, where $V' \subseteq X \times Y$ is open. Note that

where $\{X_\lambda\}$ and $\{Y_\lambda\}$ are families of open subsets of $X$ and $Y$ respectively. Then

where $O_X \subseteq X$ is open. Then

an open subset of $X$.

Now suppose $U \subseteq X$ is open. Then

which is an open set in $X \times \{y_0\}$. Thus $\tilde{f}$ is a homeomorphism, making $f$ an embedding.

We show $A = \bar{A}$. Let $x \in \bar{A}$ be a limit point of $A$. For the sake of contradiction, suppose $x \notin A$, and so $f(x) > g(x)$. If there is some $c \in Y$ such that $f(x) > c > g(x)$, then let $V_g = (-\infty, c)$ and $V_f = (c, \infty)$. Otherwise, let $V_f = (g(x), \infty)$ and $V_g = (-\infty, f(x))$. Then $V_f, V_g \subseteq Y$ are disjoint, open neighborhoods of $f(x)$ and $g(x)$, where all elements of $V_f$ are strictly greater than all elements of $V_g$. By continuity, there exist open neighborhoods $U_f, U_g \subseteq X$ of $x$ such that $f(U_f) \subseteq V_f$ and $g(U_g) \subseteq V_g$. But note $U_f \cap U_g$ is an open neighborhood of $x$. Since $x$ is a limit point, this implies there is some $u \in (U_f \cap U_g) \cap (A \setminus \{x\})$. Since $u \in A$, it follows that $f(u) \le g(u)$. But also note $f(u) \in V_f$ and $g(u) \in V_g$, meaning $f(u) > g(u)$ by how $V_f, V_g$ are defined. BY THE TRICHOTOMY POSTULATE OF ORDERED SETS, this is a contradiction. Thus $f(x) \le g(x)$, and so $x \in A$, making $A$ closed. An entirely symmetric argument shows $B = \{x: f(x) \ge g(x)\}$ is closed in $X$.

Define $\tilde{f}: A \to Y$ and $\tilde{g}: B \to Y$. By the pasting lemma, we may paste together these maps to form $h: A \cup B \to Y$ such that

which is continuous as desired.

• For our base case, suppose there are two sets in the family, $A_1$ and $A_2$. Note $f|_{A_1}$ and $f|_{A_2}$ are continuous and agree on mutual points in their domain, so gluing them together yields a continuous function, which is $f$. Now suppose the hypothesis is true for $n - 1$ many sets. Let $A' = A_1 \cup A_2 \cup \cdots \cup A_{n - 1}$, which is closed (since finite unions of closed sets are closed). $f|_{A'}$ is continuous by inductive hypothesis. Then note $f|_{A_n}$ is also continuous by the original hypothesis, and $f|_{A'}, f|_{A_n}$ agree on common points of their domain. Thus $f$ is continuous by the Pasting Lemma.

• Let $X = [0, 1]$. Let $\{A_n\}_{n \in \bZ_{\ge 0}}$ be defined such that $A_0 = \{1\}$, $A_1 = \{0\}$, and $A_n = A_{n - 1} \cup [0, 1 - 1/n]$ for every $n \ge 2 \in \bZ_+$. Then each $A_n$ is closed, since the finite union of closed intervals is closed, and $[0, 1] = \bigcup_{n \in \bZ_{\ge 0}} A_n$. Define $f: [0, 1] \to \bR$ by $f(x) = 0$ for all $x \in [0, 1)$, and $f(1) = 1$ otherwise. Then note $f|_{A_n}$ is continuous for each $n \in \bZ_+$, since $f$ is constant on this restriction. But $f$ is not continuous on $[0, 1]$, since there is a jump discontinuity at $x = 1$.

Suppose $g_1, g_2: \bar{A} \to Y$ are continuous functions such that $f = g_1|_{A} = g_2|_{A}$. For the sake of contradiction, suppose $g_1 \not \equiv g_2$, so there is some $x \in \bar{A} \setminus A$ such that $g_1(x) \ne g_2(x)$. By Hausdorffness of $Y$, draw disjoint open neighborhoods $V_1, V_2 \subseteq Y$ of $g_1(x), g_2(x)$ respectively. By continuity of $f$, note $g_1^{-1}(V_1)$ and $g_2^{-1}(V_2)$ are open subsets of $X$ that contain $x$. As such, $g_1^{-1}(V_1) \cap g_2^{-1}(V_2)$ is also an open neighborhood of $x$. Since $x$ is a limit point of $A$, there is some $x_0 \in (g_1^{-1}(V_1) \cap g_2^{-1}(V_2)) \cap (A \setminus \{x\})$. Since $x_0 \in A$, it follows that $f(x_0) = g_1(x_0) = g_2(x_0)$. But $g_1(x_0) \in V_1$ and $g_2(x_0) \in V_2$, which is a contradiction. Thus $g_1 \equiv g_2$, completing the proof.