Baby Rudin Chapter 4 Exercise Questions 5 and 6


Solution 1

For question 4.6, the forward part is not sufficient. You remark that $f(K)$ must be compact if $f$ is continuous and $K$ is compact. This is true but the graph of $f$ is not he same thing as the range of $f$. You could argue as follows: since $f(E)$ is compact, it is bounded and so $\text{graph}(f) \subset E \times f(E)$ is bounded. Since a subset of $\mathbb{R}^2$ is compact if and only if it is closed and bounded, it suffices to show that $\text{graph}(f)$ is closed. Take any convergent sequence $(x_i,f(x_i))$ in $\text{graph}(f)$. Since the canonical projections $\pi_1 : E \times f(E) \to E$ and $\pi_2 : E \times f(E) \to f(E)$ are continuous, the sequences $(x_i)$ and $f(x_i)$ are also convergent, say $x_i \to x$ and $f(x_i) \to y$. But since $f$ is continuous, $\lim_{i \to \infty} f(x_i) = f(\lim_{i \to \infty} x_i)$ so $$ (x,y) = \lim_{i\to \infty}(x_i,f(x_i)) = (x,f(x)) \in \text{graph}(f). $$ This shows that $\text{graph}(f)$ is closed, hence compact.

For the reverse direction, there are also some problems in your solution. You define $\overline{R}((u,f(u)),\epsilon)$ to be the closed ball or radius $\epsilon$ centered at $(u,f(u))$. So I guess what you mean is the intersection of this ball (as a subset of $E \times \mathbb{R}$) with $\text{graph}(f)$? But then it does not make sense later to write $f^{-1}(r)$ for $r \in \overline{R}$ as $r$ is not in the range of $f$. Also when you define $\overline{D}$, you seem to assume $f^{-1}(f(u-\epsilon))$ is a well-defined point in $\mathbb{R}$, but we don't know that the preimage of $f(u-\epsilon)$ consists of only 1 point ($f$ is not supposed injective).

Now, I am not sure I understand your idea for this but here is my attempt: The continuity of $f : E \to \mathbb{R}$ at a point $x \in E$ can be seen as the condition that $$ \lim_{a \to x} f(a) = f(x). $$ Also the compacity of $\text{graph}(f)$ here implies that for any sequence $(y_i)$ in $\text{graph}(f)$, there exists a convergence subsequence $y_{ik} \to y$ with $y \in \text{graph}(f)$. So take any sequence $(a_i)$ of $E$ converging to $x$. Then $y_i = (a_i,f(a_i))$ is a sequence in $\text{graph}(f)$ so there must exist a converging subsequence $y_{ik} = (a_{ik},f(a_{ik}))$, with $$ \lim_{k \to \infty}y_{ik} = \lim_{k \to \infty} (a_{ik},f(a_{ik})) = (x, \lim_{k\to \infty} f(a_{ik})) \in \text{graph}(f). $$ But this being in $\text{graph}(f)$ means that $f(x) = \lim_{k \to \infty}f(a_{ik})$. I'm not exactly sure how to conclude from there. It sure seems like any convergent subsequence will converge to $f(x)$ and that this is sufficient to know that $\lim_{a \to x}f(a) = f(x)$ but I'm too tired to think about this anymore :)

Solution 2

Here is a solution for 4.5, which also asks for the extension if $f:E\to R^k$.

We may assume $E$ non empty. Let $H=E^C$ then $H$ is open in $R$. We shall construct $g:R\rightarrow R^k$ continuous with $f\subset g$ - note that in the real case we could take $R$ instead of $R^1$ and the entire proof would work just fine.

If $x\in H$ we have two cases

1) $x$ is not between two elements of $E$, i.e., $x$ is an upper/lower bound for $E$. Then we map $g(x) = f(\sup E)$ (or inf depending in which side). Since $x$ is an interior point of $H$ $g$ is constant in a neighborhood of $x$ thus continuous at $x$.

2) $x$ is between two elements of $E$. Define \begin{equation*} a(x) = \sup((-\infty,x]\cap E), ~ b(x) = \inf([x,+\infty)\cap E) \end{equation*} then by $E$'s closeness both sets are closed and $a,b \in E$. We also have $(a,b) \subset H$. We construct $g$ so that its graph is a straight line between $f(a)$ and $f(b)$ in $[a,b]$: \begin{equation*} g(x) := \frac{f(b)(x-a)-f(a)(x-b)}{b-a} \end{equation*} Since for a small enough neighborhood of $x$ every component of $g$ is a straight line(i.e., a polynomial) it is continuous at $x$.

Now suppose $x\in E$. We also have two cases

1) $x$ is isolated in $E$. Then for sufficiently small $\varepsilon>0$ every $g_i$ is continuous when restricted to both $(x-\varepsilon,x]$ and $[x,x+\varepsilon)$. Now since the domain of $g$ is $R$ suffices $g(x-)=g(x)=g(x+)$ for $g$ to be continuous at $x$. We shall prove the second equality: let $(t_n)\subset (x,x+\varepsilon), t_n \to x$ then \begin{equation*} \forall i \le k, ~ \lim_{n\to\infty} g_i(t_n) = g_i(x) \end{equation*} is true since $g_i$ is a straight line in this interval. Thus \begin{equation*} \lim_{n\to\infty} g(t_n) = \left (\lim_{n\to\infty} g_i(t_n)\right) = (g_i(x)) = g(x) \end{equation*} but $t_n$ was arbitrary hence $g(x+)=g(x)$.

Again we note that this works perfectly well in $R$.

2) $x$ is a limit point of $E$: then there is at least one side of $x$ which has arbitrarily close $y\in E$, say to the right. We assert that $g(x+) = g(x)$: let $(t_n) \subset (x,x+1), t_n \to x$ then fix $\varepsilon>0$. By $f$'s continuity $\exists \delta>0$ such that \begin{equation*} y \in E, d(x,y) < \delta \Rightarrow d(g(x),g(y)) < \varepsilon \end{equation*} but such an $y$ exists exists with $x<y$ so let $N$ be such that $n\ge N$ implies \begin{equation*} x< t_n < y \end{equation*} then if $t_n \in E$ we have $d(g(t_n),g(x)) < \varepsilon$. If $t_n \in H$ then \begin{equation*} x \le a(t_n) < t_n < b(t_n) \le y \end{equation*} since $t_n$ is between two elements of $E$.

But \begin{align*} ||g(t_n)-g(x)|| = \frac{||[g(b)-g(x)](t_n-a) - [g(a)-g(x)](t_n-b)||}{b-a} \\ \le \frac{(t_n-a)d(g(x),g(b)) + (b-t_n)d(g(a),g(x))}{b-a} \\ < \frac{\varepsilon(t_n-a+b-t_n)}{b-a} = \varepsilon \end{align*} hence \begin{equation*} d(g(t_n),g(x)) < \varepsilon \end{equation*} for $n \ge N$ so \begin{equation*} \lim_{n\to\infty} g(t_n) = g(x) \end{equation*} but $t_n$ was arbitrary hence $g(x+) = g(x)$.

Now we shall end the proof with $g(x-) = g(x)$. If all left neighb. of $x$ have elements of $E$ then the previous argument works. If there exists $\varepsilon>0$ such that $(x-\varepsilon,x) \subset H$ then we invoke the proof of E1.

Therefore $g$ is continuous at all $x\in R$ and we are done.

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Updated on August 01, 2022


  • Shug
    Shug 3 months

    4.5: If f is continuous on a closed set in $R^1$, prove there exist continuous functions $g$ on $R^1$ such that $g(x)=f(x)$ for all $x \in \mathbb{E}$.

    4.6: Suppose $\mathbb{E}$ is compact, and prove that $f$ is continuous on $\mathbb{E}$ iff it's graph is compact.

    My question is two fold: With exercise 5, I just don't understand. I wrote up the proof that came to mind but obviously it's not sufficient, maybe someone can correct my thinking.

    With exercise 6, it'd be cool if someone could check over the reverse part of the iff proof and make sure that the forward portion is sufficient.

    a busy cat

    Thanks for all the help!

    • jef808
      jef808 over 8 years
      One problem with your answer in exercice 5 is that you seem to be assuming that your closed set $\mathbb{E}$ is just a closed interval. Your function doesn't really make sense if $\mathbb{E}$ is not connected or is just a bunch of points.
    • Shug
      Shug over 8 years
      Started thinking about this and it clicked. I had seen a solution before but didn't make sense and I just worked through it myself. Thanks! Hows the other proof look?
  • Shug
    Shug over 8 years
    Wow really appreciate you spending your time on this, exactly what I was looking for! A little demotivating but I'd rather be sad and right than wrong and ignorant ;)
  • jef808
    jef808 over 8 years
    No problem! Happy I can help!