You may now view your graded final exam on Gradescope. The median score was 49.5/90, the mean was 47.74/90, the standard deviation was 21.53/90, and the maximum was 82/90.
Requests for regrades must be submitted to me by email no later than Friday January 24th at 5PM. Please be precise about what and why you want regraded. Note that the primary purpose of regrades is to correct errors in grading.
Homework assignments will be available on this webpage throughout the term. All homework assignments should be submitted in class by the start of lecture on the day they are due.
Prove that $f$ is integrable, and find its integral.
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
Show that there is a metric on $\R^S$ giving rise to the topology of pointwise convergence if and only if there is a one-to-one function $\varphi : S \to \N$.
Let $R$ be the set of Riemann-integrable functions on the interval $[a, b]$. Given $f \in R$, define \[\norm{f}_2 = \paren{\int_a^b\abs{f(t)}^2\,dt}^{\frac12}.\]
The function $d_2(f, g) = \norm{f-g}_2$ on $R\times R$ is a pseudo-metric on $R$: it satisfies the triangle inequality, $d_2(f,f) = 0$, and $d_2(f, g) = d_2(g, f)$, but $d_2(f, g)=0$ does not imply $f = g$. If we define $\sim$ by $f\sim g$ whenever $d_2(f, g) = 0$, then $R/\sim$ becomes a metric space; what we have done is show that (equivalence classes of) polynomials are dense in $R/\sim$.
Let $\overline{\mathscr{A}}$ be the uniform closure $\mathscr{A}$. Show that $\overline{\mathscr{A}}$ is not an algebra. (Hint: $x\mapsto x$ is in $\overline{\mathscr{A}}$, but $x \mapsto x^2$ is not.)
Reflect on what this example says about trying to plot quadratic or linear functions on a pixelated display.
Let \begin{align*} f : \R &\longrightarrow \R \\ t &\longmapsto \begin{cases} 0 & \text{ if } t = 0 \\ t^2\sin\paren{\frac1t} & \text{ if } t \neq 0.\end{cases} \end{align*}
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
The equivalence classes $[x]_\sim \subseteq M$ are called the connected components of $M$; they are in a loose sense the "largest connected pieces" of $M$. Notice that \[M = \bigcup_{x \in M} [x]_\sim,\] that is, $M$ is the union of its connected components. Also, since $\sim$ is an equivalence relation, the connected components of $M$ are disjoint. Since every point of $M$ is in some connected component, it is easy to check that a non-empty space $M$ is connected if and only if it has precisely one connected component.
To give a few examples, the connected components of $\Z$ are the sets $\set{k}$ for each $k \in \Z$, and the connected components of $\Q$ are the sets $\set{q}$ for each $q \in \Q$. The connected components of $[0, 1) \cup \set{4} \cup (6, 9)$ are $[0, 1)$, $\set4$, and $(6,9)$. The circle $\set{(x, y) \in \R^2 \mid x^2 + y^2 = 1}$ has one connected component, itself; the same is true of any connected set. The empty set has no connected components (since in our definition, the empty set is not connected; we defined it this way so that we could unambiguously list the connected components of a set).
The equivalence classes $[x]_{\sim_p} \subseteq M$ are called path components of $M$. If $M$ has exactly one path component, it is called path connected. (Note that $\emptyset$ is not path connected: it has zero path components, not one.)
Suppose that $E \subseteq \R^n$ is open. Show that the path components of $E$ are open (in $\R^n$).
(If you find it useful, you may use without proof the fact that functions of the form \begin{align*}f : \R &\longrightarrow \R^n \\ t &\longmapsto \vec{a} + t\vec{b}\end{align*} are continuous, where $\vec{a}, \vec{b} \in \R^n$.)
It is not true that connected sets are path connected in general. Consider, for example, the set \[T = \set{(0, y) \mid -1 \leq y \leq 1} \cup \set{\paren{x, \sin\paren{\frac1x}} \mid x \gt 0} \subseteq \R^2.\] It can be shown that while $T$ is connected, the two pieces above are distinct path components of $T$.
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
Prove that $f$ is continuous at every irrational number, but discontinuous at every rational number.
Now, let $d$ and $d'$ be strongly equivalent metrics on $M$.
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
Let $M$ be a metric space, and define a metric on $M\times M$ by \begin{align*} d_2 : (M \times M) \times (M \times M) &\longrightarrow \R_{\geq0}\\ ((x_1, y_1), (x_2, y_2)) &\longmapsto \sqrt{d_M(x_1, x_2)^2 + d_M(y_1, y_2)^2}. \end{align*} (You may take for granted that this is a metric; notice that if $M = \R$, $d_2$ is the usual distance on $\R^2$.)
Prove that the original metric on $M$, $d_M : M \times M \to \R_{\geq0}$, is continuous on the metric space $(M\times M, d_2)$.
I recommend thinking about the first "Not definitions" problem below.
Suppose that $(M, d)$ and $(X, d_X)$ are metric spaces, $E \subseteq M$, $f : E \to \color{red}{X}$, and $p$ is a limit point of $E$.
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
Suppose that $M$ is a set, and let $\mathscr{F}$ be a set of pseudometrics on $M$. Let us call a subset $G \subseteq M$ open if for every $x \in M$, there are finitely many $p_1, \ldots, p_k \in \mathscr{F}$ and some $r \gt 0$ so that \[\set{y \in M \mid p_i(x, y) \lt r \text{ for each } i = 1, \ldots, k} \subseteq G.\]
We say a sequence $(a_n)_n$ in $M$ converges to $a \in M$ if for every open set $U \subseteq M$ with $a \in U$ there is some $N \in \N$ so that for every $n \gt N$, $a_n \in U$.
Due to cancelled classes, I have not had time to cover most of the material which was going to form the core of this assignment, and so the selection of problems is somewhat interesting. This assignment is optional. Any points earned will be added as bonus points to your other assignments.
Suppose that $(M, d), (X, d_X)$, and $(Y, d_Y)$ are metric spaces, and that $i_X : M \hookrightarrow X$ and $i_Y : M \hookrightarrow Y$ are isometries of $M$ into $X$ and $Y$ respectively. Let us also suppose that $X$ and $Y$ are complete: that is, that every Cauchy sequence in $X$ converges to a limit in $X$, and similar for $Y$. Finally, assume that $i_X(M) = \set{i_X(a) \mid a \in M}$ is dense in $X$ (that is, $X = \overline{i_X(M)}$).
Two metric spaces $(X, d_X)$ and $(Y, d_Y)$ are called isometric if there are isometries $I : X \hookrightarrow Y$ and $J : Y \hookrightarrow X$ which are inverses of one another. What we have now shown is that any two complete metric spaces containing $M$ as a dense manner are isometric to each other, in a way that aligns the copy of $M$ inside each. This fact is often stated as "the completion of $M$ is unique up to isometry".
Recall that earlier in the course, we merely asserted that an ordered field with the least upper bound property existed, but we didn't prove it. It is tempting to use this as our definition of what the real numbers are: to define $\R$ as $\overline{\Q}$. After making this definition, it is not so bad to show that $\R = \overline{\Q}$ has the properties of an ordered field and has the least upper bound property.
However, there is a subtle flaw with doing this. Why is it not valid to define $\R$ as $\overline{\Q}$ in this way? What could we do to avoid this problem? (No proofs needed here; just explain the problem and give a suggestion about how to avoid it.)
Show that if $K \subseteq M$ is compact, then every open double cover of $K$ has a finite open double subcover. (That is, there is a finite subset of the original open double cover which remains an open double cover.)
An earlier version of this problem had the condition "$q \gt 0$" on the set, which of course makes it fail to be closed. This has been removed.
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
Suppose that $M_1 \subseteq M_2$ and moreover that $d_1(x, y) = d_2(x, y)$ for all $x, y \in M_1$; that is, $M_1$ is a sub-metric space of $M_2$. Show that the inclusion function $\iota : M_1 \to M_2$ given by $\iota(x) = x$ is nepo.
(Note that the metric space in which a set is considered is important here! The goal is to show that open subsets of $M_2$ have preimages in $M_1$ which are open as subsets of $M_1$. You may find it useful to first show that $\iota^{-1}(U) = U \cap M_2$.)
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
Show that if $d$ is an anti-metric on $S$, then $S$ contains at most one element.
Suppose that $\emptyset \subsetneq E_1 \subseteq E_2 \subseteq \R$. If $E_2$ is bounded above, then $\sup E_1$ exists and $\sup E_1 \leq \sup E_2$.
(Note: the symbol "$\subsetneq$" means "is a proper subset of", so $A \subsetneq B$ means $A \subseteq B$ and $A \neq B$. Thus $\emptyset \subsetneq E_1$ means "$E_1$ is non-empty". In contrast, the symbol $\nsubseteq$ means "is not a subset of".)
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
Prove that if $\preceq$ is a partial order on $S$, there is an order on $S$ which extends $\preceq$: that is, an order $\preceq'$ on $S$ so that if $x, y \in S$ with $x \preceq y$, then $x \preceq' y$.
To prove this, you may assume the following:
Zorn's Lemma. Suppose that $Z$ is a set and $\preceq$ is a partial order on $Z$ with the property that every totally-ordered subset of $Z$ is bounded above in $Z$. Then $Z$ contains a maximal element. (A subset $W \subseteq Z$ is totally ordered if $\preceq$ restricted to $W$ is an order, i.e., if whenever $x, y \in W$ we have either $x \preceq y$ or $y \preceq x$. An element $x \in Z$ is maximal if whenever $y \in Z$ with $x \preceq y$, we have $y = x$.)
You may wish to type your homework; for example, this makes it much easier for others to read, and makes it easier to edit and produce a coherent final argument. Most modern mathematics papers are typeset using a system called \(\mathrm{\LaTeX}\) (pronounced "lah-tech" or "lay-tech"; see the Wikipedia entry on Pronouncing and writing "LaTeX"). Although it has a steep learning curve, it is extremely useful for typesetting complicated mathematical expressions. There are many resources available online, such as this reference by Oetiker, Partl, Hyna, and Schlegl. I have also made available an assignment template here, which produces this output when compiled correctly.
Here is a practice midterm. It is roughly 25% longer than the midterm. It may be slightly more difficult, too.
On problem 4.B. of the practice midterm, the assumptions on $g$ were meant to be "$g$ is continuous at $0$ and $g(0) \gt 0$"; the problem is in fact much easier with the original assumptions. As further practice, try solving it under these weaker assumptions.
Lemma (The Archimedean Property): $\Z$ is not bounded above in $\R$.
Proof: Suppose that $\Z$ were bounded above. Since it is non-empty, by the Least Upper Bound Property of $\R$ it has a least upper bound $w = \sup\Z$. Now by Homework 1 Problem 3 C, $\sup\set{k +1 \mid k \in \Z} = w + 1$. But $\set{k + 1 \mid k \in \Z} = \Z$, so $w = \sup\Z = w + 1$. Adding $-w$ to both sides, we find $0 = (-w) + w = (-w) + w + 1 = 0 + 1 = 1$, contradicting that $\R$ is a field (and so $0 \neq 1$).
Theorem: Suppose $\alpha, \beta \in \R$ with $\alpha \lt \beta$. Then there is some $q \in \Q$ with $\alpha \lt q \lt \beta$.
Proof: Since $\alpha \lt \beta$ we have $\beta - \alpha \gt 0$, so $(\beta - \alpha)^{-1} \gt 0$. Now $(\beta - \alpha)^{-1}$ is not an upper bound for $\Z$ (since $\Z$ is not bounded above in $\R$) and so there is some $k \in \Z$ with $k \gt (\beta - \alpha)^{-1}$. Then in particular, $k^{-1} \lt \beta - \alpha$.
Let $T = \set{n \in \Z \mid nk^{-1} \lt \beta}$. Notice first $T$ is bounded above (by $k\beta$). Moreover, $T$ is not empty: indeed, if $T$ were empty, it would mean that $n \geq k\beta$ for every $n \in \Z$. But that would mean that $-n \leq -k\beta$ for every $n \in \Z$, so $-k\beta$ is an upper bound for $\set{-n \mid n \in \Z} = \Z$, contradicting the lemma above. Thus $T$ is non-empty.
Now let $t \in T$, and write $T_+ = \set{n \in \Z \mid nk^{-1} \lt \beta, n \geq t}$. If $s \in \Z$ is such that $s \gt k\beta$, then $T_+ \subseteq \set{t, t+1, \ldots, s}$ which is a finite set. Therefore $T_+$ is a non-empty finite set (for $t \in T_+$), and $\sup T_+ \in T_+ \subset \Z$. Let us write $q = k^{-1}\sup T_+$, so that $q \lt \beta$ and $q \in \Q$; if we can show $q \gt \alpha$, we will be done.
First, observe that $q + k^{-1} \geq \beta$, for otherwise, $kq \lt kq + 1 \in T_+$, contradicting that $kq = \sup T_+$. But then $\beta - q \leq k^{-1} \lt \beta - \alpha$. Adding $-\beta$ to both sides, we find $-q \lt -\alpha$; adding $q + \alpha$ to both sides then yeilds $\alpha \lt q$, as desired.
The intuition in the above proof is this: if $k^{-1} \lt \beta - \alpha$, some multiple of $k^{-1}$ must fall between $\alpha$ and $\beta$. If you cover the real line with a mesh with gaps of size smaller than the space between two numbers, something must land in that space.
The reason we drop to the set $T_+$ in the third paragraph is to ensure that our candidate number is actually rational. While it is true that the supremum of any set of integers bounded above is again an integer, I did not want to go through that argument here; it is easier to truncate to a finite set and use the homework.