Homework 3

Solution.

(iii) $\implies$ (i):

Let $x_0 \in X$ and $\varepsilon > 0$. Consider $B_Y\p{f\p{x_0}, \varepsilon}$, which is an open set (by the long proposition). Thus, by assumption, $f^{-1}\p{B_Y\p{f\p{x_0}, \varepsilon}}$ is also open and contains $x_0$ by construction. Thus, (by the long proposition again, $x_0$ is an interior point, so) there exists $\delta > 0$ such that $B_X\p{x_0, \delta} \subseteq f^{-1}\p{B_Y\p{f\p{x_0}, \varepsilon}}$. In other words,

so $f$ is continuous at $x_0$ because $\varepsilon$ was arbitrary. Since $x_0$ was also arbitrary, this shows that $f$ is continuous.

(iii) $\iff$ (iv):

Given any subset $E \subseteq Y$, we have

In other words,

If (iii) holds, then given a closed set $F \subseteq Y$, (the long proposition implies) its complement $Y \setminus F$ is open, so by (iii), $f^{-1}\p{Y \setminus F}$ is open and by ($*$), its complement is $f^{-1}\p{F}$ and must be closed.

Similarly, if (iv) holds, then given an open $V \subseteq Y$, its complement $Y \setminus V$ is closed. Using (iv), $f^{-1}\p{Y \setminus V}$ is closed and using ($*$) again, we see $f^{-1}\p{V}$ must be open.

Examples

1. Let $X = Y = \R$ and consider the constant function $f\p{x} = 0$, which is trivially continuous. Then $U = X$ is open (since the entire metric space is both open and closed in itself), but $f\p{U} = \set{0}$, which is not open in $\R$.

2. Let $X = Y = \R$ and consider the function $f\p{x} = \frac{1}{1 + x^2}$, which is continuous since $1 + x^2$ is continuous and is never $0$. If we take $K = \R$, then $K$ is closed (by the same reasoning above) and $f\p{K} = \poc{0, 1}$, which is not closed in $\R$.

Common Mistakes

The most common mistakes were giving examples since the choice of $X$ and $Y$ can make or break your example. For example, if you define $\func{f}{\R}{\set{0}}$ by $f\p{x} = 0$, then while it's true that $f\p{\R} = \set{0}$ still, the problem is that in the (discrete) metric space $\set{0}$, the singleton $\set{0}$ is open.

Similarly, I saw some students use $\func{f}{\R}{\pco{0, \infty}}$, $f\p{x} = x^2$, and $U = \p{-1, 1}$. The issue here is that $\pco{0, \infty}$ has different open sets from $\R$ itself. Indeed, $f\p{U} = \poc{0, 1}$ which is not open in $\R$, but is relatively open in $\pco{0, \infty}$ since

See Proposition 1.3.4.

Solution.

First, if we can show that all continuous functions on compact metric spaces attain a maximum, then they also attain a minimum. Indeed, note that $-f$ is continuous, so there exists $x_{\min} \in X$ such that $-f\p{x_{\min}} = \sup \p{-f\p{X}}$ by assumption. Finally, recall from 131A that

so $f$ also attains a minimum. Thus, we only need to show that $f$ attains its maximum. (Alternatively, you can just redo the whole proof with $\sup f\p{X}$ replaced with $\inf f\p{X}$.)

Since $X$ is compact and $f$ is continuous, its image $f\p{X}$ is also compact, which means $f$ is automatically bounded. If $X \neq \emptyset$, then $f\p{X} \neq \emptyset$ and is bounded, so $\sup f\p{X}$ exists in $\R$. Recall (from 131A) that there is a sequence $\p{y_n}_n \subseteq f\p{X}$ such that $y_n \to \sup f\p{X}$.

From here, there are two ways to finish the problem.

Solution 1 (using abstract nonsense)

We have shown that $\sup f\p{X}$ is an adherent point of $f\p{X}$. Since $f\p{X}$ is compact, it is also closed, so $\sup f\p{X} \in f\p{X}$ or in other words, there exists $x_{\max} \in X$ such that $f\p{x_{\max}} = \sup f\p{X}$.

Solution 2 (using sequences)

By definition of $f\p{X}$, for each $n \in \N$, there exists $x_n \in X$ such that $f\p{x_n} = y_n$. Since $X$ is compact, there is a convergent subsequence $\p{x_{n_k}}_k$ which converges to some $x_0 \in X$. Thus, because $f$ is continuous,

so we can take $x_{\max} = x_0$.