#2. Suppose $f: X \rightarrow Y$ is continuous. If $x$ is a limit point of the subset $A$ of $X$, is it necessarily true that $f(x)$ is a limit point of $f(A)$?

If $f: X \rightarrow Y$ is continuous and $x$ is a limit point of $A \subset X$, it is not necessarily true that $f(x)$ is a limit point of $f(A)$. We prove this by giving a counterexample.

Consider the sets $X = \{a, b\}$ and $Y = \{c\}$, each with the trivial topology, and let $f(a) = f(b) = c$. Clearly $f$ is continuous because the pre-image of the only open sets, namely $Y$ and $\varnothing$, are $X$ and $\varnothing$ respectively, which are open.

Then define $A = \{a\}$. Clearly $b$ is a limit point of $A$, because every neighborhood of $b$ (which in this case is only $X$ itself) intersects $A - \{b\}$. Note that $f(b) = c$ and $f(A) = Y$. But $c$ is not a limit point of $Y$, so $f(b)$ is not a limit point of $f(A)$.

#8. Let $Y$ be an ordered set in the order topology, and let $f, g: X \rightarrow Y$ be continuous.

(a) Show that $C = \{ x \mid f(x) \leq g(x) \}$ is closed.

We show that $C$ is closed by proving that if $x \notin C$, then $x$ is not a limit point of $C$.

Suppose $x_0 \notin C$. Then $f(x_0) > g(x_0)$. Note that the order topology is Hausdorff (per §17 #10). Then there exist disjoint neighborhoods $U$ and $V$ of $f(x_0)$ and $g(x_0)$ respectively. Then there exist basis elements $B_1$ and $B_2$, i.e. intervals in the order relation, such that $f(x_0) \in B_1 \subset U$ and $g(x_0) \in B_2 \subset V$. Then since $f(x_0) > g(x_0)$ and by the disjointness of the intervals, all elements of $B_1$ are larger than all elements of $B_2$.

Since $B_1$ and $B_2$ are open, and $f$ and $g$ are continuous, $f^{-1}(B_1)$ and $g^{-1}(B_2)$ are open. Since intersections of open sets are open, $f^{-1}(B_1) \cap g^{-1}(B_2)$ is open, and $x_0$ is in this set. Thus we have an open neighborhood of $x_0$ in which $f(x) > g(x)$ for all $x$ in the neighborhood. Thus $x_0$ has an open neighborhood disjoint from $C$, so $x$ is not a limit point of $C$.

Thus $C$ contains all its limit points, so by corollary 17.7, $C$ is closed.

(b) Let $h: X \rightarrow Y$ be the function $h(x) = \text{min}\{ f(x), g(x) \}$. Show that $h$ is continuous.

Let $A = \{ x \mid f(x) \leq g(x) \}$ and $B = \{ x \mid g(x) \leq f(x) \}$. Clearly $X = A \cup B$ and per part (a), both $A$ and $B$ are closed. Also, $f$ and $g$ must agree on $A \cap B$, because if $x \in A \cap B$ then $f(x) \leq g(x) \leq f(x)$. Finally, because the restriction of continuous functions is continuous, the functions $f|_A$ and $g|_B$ are continuous. Then the minimizing function $h$ is just the combination of $f|_A$ and $g|_B$ as in the pasting lemma, so it is also continuous.

#9. Let $\{ A_\alpha \}$ be a collection of subsets of $X$; let $X = \bigcup_\alpha A_\alpha$. Let $f: X \rightarrow Y$; suppose that $f \vert A_\alpha$ is continuous for each $\alpha$.

(a) Show that if the collection $\{ A_\alpha \}$ is finite and each set $A_\alpha$ is closed, then $f$ is continuous.

Given that the collection is finite, we can apply the pasting lemma iteratively. So apply it on $A_1$ and $A_2$ to get a continuous function on $A_1 \cup A_2$, then apply it on this set and $A_3$ to get a continuous function on $A_1 \cup A_2 \cup A_3$, and so on.

(b) Find an example where the collection $\{ A_\alpha \}$ is countable and each $A_\alpha$ is closed, but $f$ is not continuous.

Had trouble with this. Some thoughts.

(c) An indexed family of sets $\{ A_\alpha \}$ is said to be locally finite if each point $x$ of $X$ has a neighborhood that intersects $A_\alpha$ for only finitely many values of $\alpha$. Show that if the family $\{ A_\alpha \}$ is locally finite and each $A_\alpha$ is closed, then $f$ is continuous.

Note that for each $x \in X$, there is a neighborhood $U_x$ of $x$ that only intersects with finitely many $A_\alpha$. So for each $x$ there is some collection $A_{\alpha_1}, \ldots, A_{\alpha_N}$ and a neighborhood $U_x$ such that

$$ x \in U_x \subset A_{\alpha_1} \cup \ldots \cup A_{\alpha_N} $$

Now note that by the pasting lemma and part (a), $f$ is continuous on $A_{\alpha_1} \cup \ldots \cup A_{\alpha_N}$. Then, $f$ is continuous on $U_x$. (I'm actually not 100% sure about restricting down to $U_x$ but I think it works.)

Finally, by the local formulation of continuity (theorem 18.2), $f$ is continuous on $X$.

#13. Let $A\subset X$; let $f: A \rightarrow Y$ be continuous; let $Y$ be Hausdorff. Show that if $f$ may be extended to a continuous function $g: \overline{A} \rightarrow Y$, then $g$ is uniquely determined by $f$.

Let $g_1, g_2: \overline{A} \rightarrow Y$ be arbitrary continuous functions such that for $x \in A$, $g_1(x) = g_2(x) = f(x)$. We show that $g_1$ and $g_2$ are equal on $\overline{A}$.

Suppose $x_0 \in \overline{A}$. Let us assume for contradiction that $g_1(x_0) \neq g_2(x_0)$. Then since $Y$ is Hausdorff, there exist open neighborhoods $U$ of $g_1(x_0)$ and $V$ of $g_2(x_0)$ respectively that are disjoint. And since $g_1$ and $g_2$ are continuous on $\overline{A}$, the sets $g_1^{-1} (U)$ and $g_2^{-1}(V)$ are open in $\overline{A}$. By definition of the subspace topology, these sets are equal to $W_1 \cap \overline{A}$ and $W_2 \cap \overline{A}$ for some $W_1$, $W_2$ open in $X$.

By construction, $x_0$ is in both of these sets:

$x_0 \in g_1^{-1}(U) \cap g_2^{-1} (V) = (W_1 \cap \overline{A}) \cap (W_2 \cap \overline{A})$

Thus $W_1 \cap W_2$ is an open neighborhood of $x_0$ in $X$. Then since $x_0 \in \overline{A}$, there exists some $p \in W_1 \cap W_2$ such that $p \in A$. So we have

$p \in A \cap W_1 \cap W_2 \implies p \in g_1^{-1}(U) \cap g_2^{-1}(V)$

So $g_1(p) \in U$ and $g_2(p) \in V$. But since $p$ is in $A$, $g_1$ and $g_2$ must agree on it, so $g_1(p) = g_2(p)$. Then $U$ and $V$ are not disjoint, so we have a contradiction.

Thus $g_1$ and $g_2$ are equal on $\overline{A}$, so $g$ is uniquely determined by $f$.