# Abstract Nonsense

## Munkres Chapter 2 Section 19 (Part I)

1.

Problem: Suppose that for each $\alpha\in\mathcal{A}$ the topology on $X_\alpha$ is given by a basis $\mathfrak{B}_\alpha$. The collection of all sets of the form

$\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha$

Such that $B_\alpha\in\mathfrak{B}_\alpha$ is a basis for $\displaystyle X=\prod_{\alpha\in\mathcal{A}}X_\alpha$ with the box topology, denote this collection by $\Omega_B$. Also, the collection $\Omega_P$ of all sets of the form

$\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha$

Where $B_\alpha\in\mathfrak{B}_\alpha$ for finitely many $\alpha$ and $B_\alpha=X_\alpha$ otherwise is a basis for the product topology on $X$.

Proof: To prove the first part we let $U\subseteq X$ be open. Then, by construction of the box topology for each $(x_\alpha)_{\alpha\in\mathcal{A}}\overset{\text{def.}}{=}(x_\alpha)\in U$ we may find some $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha$ such that $U_\alpha$ is open in $X_\alpha$ and $\displaystyle (x_\alpha)\in \prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U$. So, then for each $x_\alpha$ we may find some $B_\alpha\in\mathfrak{B}_\alpha$ such that $x_\alpha\in B_\alpha\subseteq U_\alpha$ and thus

$\displaystyle (x_\alpha)\in\prod_{\alpha\in\mathcal{A}}B_\alpha\subseteq\prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U$

Noticing that $\displaystyle \prod_{\alpha\in\mathcal{A}}B_\alpha\in\Omega_B$ and every element of $\Omega_B$ is open finishes the argument.

Next, we let $U\subseteq X$ be open with respect to the product topology. Once again for each $(x_\alpha)\in U$ we may find some $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha$ such that $U_\alpha$ is open in $X_\alpha$ for each $\alpha\in\mathcal{A}$ and $U_\alpha=X_\alpha$ for all but finitely many $\alpha$, call them $\alpha_1,\cdots,\alpha_m$. So, for each $\alpha_k,\text{ }k=1,\cdots,m$ we may find some $B_k\in\mathfrak{B}_k$ such that $x\in B_k\subseteq U_k$ and so

$\displaystyle (x_\alpha)\in\prod_{\alpha\in\mathcal{A}}V_\alpha\subseteq\prod_{\alpha\in\mathcal{A}}U_\alpha\subseteq U$

Where

$\displaystyle V_\alpha=\begin{cases}B_k\quad\text{if}\quad \alpha=\alpha_k ,\text{ }k=1,\cdots,m\\ X_\alpha\quad\text{if}\quad x\ne \alpha_1,\cdots,\alpha_m\end{cases}$

Noting that $\displaystyle \prod_{\alpha\in\mathcal{A}}V_\alpha\in\Omega_P$ and $\Omega_P$ is a collection of open subsets of $X$ finishes the argument. $\blacksquare$

2.

Problem: Let $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of topological spaces such that $U_\alpha$ is a subspace of $X_\alpha$ for each $\alpha\in\mathcal{A}$. Then, $\displaystyle \prod_{\alpha\in\mathcal{A}}U_\alpha=Y$ is a subspace of $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=X$ if both are given the product or box topology.

Proof: Let $\mathfrak{J}_S,\mathfrak{J}_P$ denote the topologies $Y$ inherits as a subspace of $X$ and as a product space respectively. Note that $\mathfrak{J}_S,\mathfrak{J}_P$ are generated by the bases $\mathfrak{B}_S=\left\{B\cap Y:B\in\mathfrak{B}\right\}$ (where $\mathfrak{B}$ is the basis on $X$ with the product topology), and

$\displaystyle \mathfrak{B}_P=\left\{\prod_{\alpha\in\mathcal{A}}O_\alpha:O_\alpha\text{ is open in }U_\alpha,\text{ and equals }U_\alpha\text{ for all but finitely many }\alpha\right\}$

So, let $(x_\alpha)\in B$ where $B\in\mathfrak{B}_S$ then

$\displaystyle B=Y\cap\prod_{\alpha\in\mathcal{A}}V_\alpha=\prod_{\alpha\in\mathcal{A}}\left(V_\alpha\cap U_\alpha\right)$

Where $V_\alpha$ is open in $X_\alpha$, and thus $V_\alpha\cap Y$ is open in $U_\alpha$. Also, since $V_\alpha=X_\alpha$ for all but finitely many $\alpha$ it follows that $V_\alpha\cap U_\alpha=U_\alpha$ for all but finitely many $\alpha$. And so $B\in\mathfrak{B}_P$. Similarly, if $B\in\mathfrak{B}_P$ then

$\displaystyle B=\prod_{\alpha\in\mathcal{A}}O_\alpha$

Where $O_\alpha$ is open in $U_\alpha$, but this means that $O_\alpha=V_\alpha\cap U_\alpha$ for some open set $V_\alpha$ in $X_\alpha$ and so

$\displaystyle B=\prod_{\alpha\in\mathcal{A}}O_\alpha=\prod_{\alpha\in\mathcal{A}}\left(V_\alpha\cap U_\alpha\right)=\prod_{\alpha\in\mathcal{A}}V_\alpha\cap Y\in\mathfrak{B}_S$

From where it follows that $\mathfrak{B}_S,\mathfrak{B}_P$ and thus $\mathfrak{J}_S,\mathfrak{J}_P$ are equal.

The case for the box topology is completely analgous. $\blacksquare$

3.

Problem: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of Hausdorff spaces, then $\displaystyle X=\prod_{\alpha\in\mathcal{A}}X_\alpha$ is Hausdorff with either the box or product topologies

Proof: It suffices to prove this for the product topology since the box topology is finer.

So, let $(x_\alpha),(y_\alpha)\in X$ be distinct. Then, $x_\beta\ne y_\beta$ for some $\beta\in\mathcal{A}$. Now, since $X_\beta$ is Hausdorff there exists disjoint open neighborhoods $U,V$ of $x_\beta,y_\beta$ respectively. So, $\pi_\beta^{-1}(U),\pi_\beta^{-1}(V)$ are disjoint open neighborhoods of $(x_\alpha),(y_\alpha)$ respectively. The conclusion follows. $\blacksquare$

4.

Problem: Prove that $\left(X_1\times\cdots\times X_{n-1}\right)\times X_n\overset{\text{def.}}{=}X\approx X_1\times\cdots\times X_n\overset{\text{def.}}{=}Y$.

Proof: Define

$\displaystyle \varphi:X\to Y:((x_1,\cdots,x_{n-1}),x_n)\mapsto (x_1,\cdots,x_n)$

Clearly this is continuous since $\pi_{\beta}\circ\varphi=\pi_\beta$

5.

Problem: One of the implications holds for theorem 19.6 even in the box topology, which is it?

Proof: If $\displaystyle f:A\to\prod_{\alpha\in\mathcal{A}}X_\alpha$ where the latter is given the box topology then we have that each $\pi_\alpha$ is continuous and thus so is each $\pi_\alpha\circ f:A\to X_\alpha$. $\blacksquare$#

6.

Problem: Let $\left\{\bold{x}_n\right\}_{n\in\mathbb{N}}$ be a sequence of points in the product space $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha=X$. Prove that $\left\{\bold{x}_n\right\}_{n\in\mathbb{N}}$ converges to $\bold{x}$ if and only if the sequences $\left\{\pi_\alpha(\bold{x}_n)\right\}_{n\in\mathbb{N}}$ coverge to $\pi_\alpha(\bold{x})$ for each $\alpha\in\mathcal{A}$. Is this fact true if one uses the box topology?

Proof: Suppose that $U$ is a neighborhood of $\pi_{\alpha}(\bold{x})$ such that

$\left(X_\alpha-U\right)\cap\left\{\bold{x}_n:n\in\mathbb{N}\right\}=K$

is infinite. Notice then that if $\pi_{\alpha}(\bold{x}_n)\in K$ that $\bold{x}_n\notin\pi_{\alpha}^{-1}\left(U\right)$ from where it follows that $\pi_{\alpha}^{-1}\left(U\right)$ is a neighborhood of $\bold{x}$ which does not contain all but finitely many values of $\left\{\bold{x}_n:n\in\mathbb{N}\right\}$ contradicting the fact that $\bold{x}_n\to\bold{x}$ in $X$.

Conversely, suppose that $\pi_{\alpha}(\bold{x}_n)\to\pi_{\alpha}(\bold{x})$ for each $\alpha\in\mathcal{A}$  and let $\displaystyle \prod_{\alpha\in\mathcal{A}}U_{\alpha}$ be a basic open neighborhood of $\bold{x}$. Then, letting $\alpha_1,\cdots,\alpha_m$ be the finitely many indices such that $U_{\alpha_k}\ne X_k$. Since each $\pi_{\alpha_k}(\bold{x}_n)\to\pi_{\alpha_k}(\bold{x})$ there exists some $n_\ell\in\mathbb{N}$ such that $n_\ell\leqslant n\implies \pi_{\alpha_k}(\bold{x}_k)\in U_k$. So, let $N=\max\{n_1,\cdots,n_k\}$. Now, note that if $\displaystyle \bold{x}_n\notin\prod_{\alpha\in\mathcal{A}}U_\alpha$ then $\pi_{\alpha}(\bold{x}_n)\notin U_\alpha$ for some $\alpha\in\mathcal{A}$. But, since clearly $\pi_{\alpha}(\bold{x}_n)\in X_\alpha$ we must have that $\pi_{\alpha}(\bold{x}_n)\notin U_{\alpha_1}\cup\cdots\cup U_{\alpha_m}$ and thus $n\leqslant N$. It follows that for every $N\leqslant n$ we have that $\displaystyle \bold{x}_n\in\prod_{\alpha\in\mathcal{A}}U_\alpha$. Then, since every neighborhood of $\bold{x}$ contains a basic open neighborhood and the above basic open neighborhood was arbitrary the conclusion follows.

Clearly the first part holds in the box topology since there was no explicit mention of the product topology or it’s defining characteristics. That said, the second does not hold. Consider $\displaystyle \bold{x}_n=\prod_{m\in\mathbb{N}}\left\{\frac{1}{n}\right\}$. Clearly each coordinate converges to zero, but $\displaystyle \prod_{m\in\mathbb{N}}\left(\frac{-1}{m},\frac{1}{m}\right)=U$ is a neighborhood of $\bold{0}$ in the product topology. But, if one claimed that for every $n\geqslant N$ (for some $N\in\mathbb{N}$ that $\bold{x}_n\in U$ they’d be wrong. To see this merely note $\displaystyle \frac{1}{N}\notin\left(\frac{-1}{N+1},\frac{1}{N+1}\right)$ and so $\pi_{N+1}\left(\bold{x}_N\right)\notin\pi_{N+1}(U)$ and thus $\bold{x}_{N}\notin U$.

7.

Problem: Let $\mathbb{R}^{\infty}$ be the subset of $\mathbb{R}^{\omega}$ consisting of all eventually zero sequences. What is $\overline{\mathbb{R}^{\infty}}$ in the box and product topology?

Proof: We claim that in the product topology $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\omega}$. To see this let $\displaystyle \prod_{n\in\mathbb{N}}U_n$ be a basic non-empty open set in $\mathbb{R}^{\omega}$ with the product topology. Since we are working with the product topology we know there are finitely many indices $n_1,\cdots,n_m$ such that $U_{n_k}\ne \mathbb{R}$. So, for each $n_1,\cdots,n_m$ select some $x_{n_k}\in U_{n_k}$ and consider $(x_n)_{n\in\mathbb{N}}$ where

$\displaystyle \begin{cases} x_{n_k}\quad\text{if}\quad n=n_1,\cdots,n_m \\ 0 \text{ } \quad\text{if}\quad \text{otherwise}\end{cases}$

Clearly then $\displaystyle (x_n)_{n\in\mathbb{N}}\in \prod_{n\in\mathbb{N}}U_n\cap\mathbb{R}^{\infty}$ and thus every non-empty open set in $\mathbb{R}^{\omega}$ intersects $\mathbb{R}^{\infty}$ and the conclusion follows.

Now, we claim that with the box topology that $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}$. To see this let $(x_n)_{n\in\mathbb{N}}\notin\mathbb{R}^{\infty}$. Then, there exists some subsequence $\{x_{\varphi(n)}\}$ of the sequence $\{x_n\}$ which is non-zero. For each $\varphi(n)$ form an interval $I_{\varphi(n)}$ such that $0\notin I_{\varphi(n)}$. Then, consider

$\displaystyle \prod_{n\in\mathbb{N}}U_n,\text{ }U_n=\begin{cases} I_{\varphi(m)}\quad\text{if}\quad n=\varphi(m)\\ \mathbb{R}\quad\text{ }\quad\text{if}\quad \text{otherwise}\end{cases}$

Clearly then $\displaystyle \prod_{n\in\mathbb{N}}U_n$ is a neighborhood of $(x_n)_{n\in\mathbb{N}}$ and since each clearly has an infinite subsequence of non-zero values it is disjoint from $\mathbb{R}^{\infty}$. It follows that in $\mathbb{R}^{\omega}$ with the box topology that $\mathbb{R}^{\infty}$ is closed and thus $\overline{\mathbb{R}^{\infty}}=\mathbb{R}^{\infty}$ as desired. $\blacksquare$

8.

Problem: Given sequences $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ of real numbers with $a_n>0$ define $\varphi:\mathbb{R}^{\omega}\to\mathbb{R}^{\omega}$ by the equation

$\varphi:(x_n)_{n\in\mathbb{N}}\mapsto \left(a_n x_n+b_n\right)_{n\in\mathbb{N}}$

Show that if $\mathbb{R}^{\omega}$ is given the product topology that $\varphi$ is a homeomorphism. What happens if $\mathbb{R}^{\omega}$ is given the box topology?

Proof: Let us first prove that $\varphi$ is a bijection. To do this we prove something more general…

Lemma: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}}$ be two classes of untopologized sets and $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ a collection of bijections $f_\alpha:X_\alpha\to Y_\alpha$. Then, if

$\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}$

we have that $\varphi$ is a bijection.

Proof: To prove injectivity we note that if

$\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\varphi\left((y_\alpha)_{\alpha\in\mathcal{A}}\right)$

Then,

$\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}=\left(f_\alpha(y_\alpha)\right)_{\alpha\in\mathcal{A}}$

And by definition of an $\alpha$-tuple this implies that

$f_\alpha(x_\alpha)=f_\alpha(y_\alpha)$

for each $\alpha\in\mathcal{A}$. But, since each $f_\alpha:X_\alpha\to Y_\alpha$ is injective it follows that

$x_\alpha=y_\alpha$

For each $\alpha\in\mathcal{A}$. Thus,

$(x_\alpha)_{\alpha\in\mathcal{A}}=(y_\alpha)_{\alpha\in\mathcal{A}}$

as desired.

To prove surjectivity we let $\displaystyle (y_\alpha)_{\alpha\in\mathcal{A}}\in\prod_{\alpha\in\mathcal{A}}Y_\alpha$ be arbitrary. We then note that for each fixed $\alpha\in\mathcal{A}$ we have there is some $x_\alpha\in X_\alpha$ such that $f_\alpha(x_\alpha)=y_\alpha$. So, if $\displaystyle (x_\alpha)_{\alpha\in\mathcal{A}}\in\prod_{\alpha\in\mathcal{A}}X_\alpha$ is the corresponding $\alpha$-tuple of these values we have that

$\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}=(y_\alpha)_{\alpha\in\mathcal{A}}$

from where surjectivity follows. Combining these two shows that $\varphi$ is indeed a bijection. $\blacksquare$

Lemma: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}}$ be two classes of non-empty topological spaces and $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ a corresponding class of continuous functions such that $f_\alpha:X_\alpha\to Y_\alpha$. Then, if $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ and $\displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha$ are given the product topologies the mapping

$\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}$

is continuous.

Proof: Since the codomain is a product space it suffices to show that

$\displaystyle \pi_\beta\circ\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to Y_{\beta}$

is continuous for each $\beta\in\mathcal{A}$. We claim though that the diagram

$\begin{matrix}&space;\prod_{\alpha\in\mathcal{A}}X_{\alpha}&space;&\xrightarrow[\quad\quad\quad]{\pi^X_\beta}&space;&X_\beta&space;\\&space;^{\varphi}\bigg\downarrow&&space;&\bigg\downarrow{^{f_\beta}}&space;\\&space;\prod_{\alpha\in\mathcal{A}}Y_\alpha&\xrightarrow[\quad\quad\quad]{\pi^Y_\beta}&space;&&space;Y_\beta&space;\end{matrix}$

commutes where $\pi^Y_\beta$ and $\pi^X_\beta$ denote the canonical projections from $\displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha$ and $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ to $Y_\beta$ and $X_\beta$ respectively. To see this we merely note that

$\displaystyle \pi^Y_\beta\left(\varphi\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)\right)=\pi^Y_\beta\left(\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}\right)=f_{\beta}(x_\beta)$

and

$f_\beta\left(\pi^X_\beta\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)\right)=f_\beta\left(x_\beta\right)$

which confirms the commutativity of the diagram. But, the conclusion follows since $f_\beta\circ\pi_\beta$ is the composition of two continuous maps (the projection being continuous since $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ is a product space).

The lemma follows by previous comment. $\blacksquare$

We come to our last lemma before the actual conclusion of the problem.

Lemma: Let $\left\{X_\alpha\right\}_{\alpha\in\mathcal{A}}$ and $\left\{Y_\alpha\right\}_{\alpha\in\mathcal{A}}$ be two classes of non-empty topological spaces and $\left\{f_\alpha\right\}_{\alpha\in\mathcal{A}}$ a set of homeomorphisms with $f_\alpha:X_\alpha\to Y_\alpha$. Then,

$\displaystyle \prod_{\alpha\in\mathcal{A}}f_\alpha\overset{\text{def.}}{=}\varphi:\prod_{\alpha\in\mathcal{A}}X_\alpha\to\prod_{\alpha\in\mathcal{A}}Y_\alpha:(x_\alpha)_{\alpha\in\mathcal{A}}\mapsto\left(f_\alpha(x_\alpha)\right)_{\alpha\in\mathcal{A}}$

is a homeomorphism if $\displaystyle \prod_{\alpha\in\mathcal{A}}X_\alpha$ and $\displaystyle \prod_{\alpha\in\mathcal{A}}Y_\alpha$ are given the product topology.

Proof: Our last two lemmas show that $\varphi$ is bijective and continuous. To prove that it’s inverse is continuous we note that

$\displaystyle \left(\left(\prod_{\alpha\in\mathcal{A}}f_{\alpha}^{-1}\right)\circ\varphi\right)\left((x_\alpha)_{\alpha\in\mathcal{A}}\right)=\prod_{\alpha\in\mathcal{A}}\left\{f_\alpha^{-1}\left(f_\alpha(x_\alpha)\right)\right\}=(x_\alpha)_{\alpha\in\mathcal{A}}$

And similarly for the other side. Thus,

$\displaystyle \varphi^{-1}=\prod_{\alpha\in\mathcal{A}}f_{\alpha}^{-1}$

Which is continuous since each $f_{\alpha}^{-1}:Y_\alpha\to X_\alpha$ is continuous and appealing to our last lemma again. Thus, $\varphi$ is a bi-continuous bijection, or in other words a homeomorphism. The conclusion follows. $\blacksquare$

Thus, getting back to the actual problem we note that if we denote $T_n:\mathbb{R}\to\mathbb{R}:x\mapsto a_nx+b_n$ that each $T_n$ is a homeomorphism. Thus, since it is easy to see that

$\displaystyle \varphi=\prod_{n\in\mathbb{N}}T_n$

we may conclude by our last lemma (since we are assuming that we are giving $\mathbb{R}^{\omega}$ in both the domain and codomain the product topology) that $\varphi$ is a homeomorphism.

This is also continuous if we give $\mathbb{R}^{\omega}$ the box topology. To see this we merely need to note that $\displaystyle \left(\prod_{\alpha\in\mathcal{A}}f_\alpha\right)^{-1}\left(\prod_{\alpha\in\mathcal{A}}U_\alpha\right)=\prod_{\alpha\in\mathcal{A}}f_\alpha^{-1}\left(U_\alpha\right)$ and thus if all of the $U_\alpha$ are open then so are (since each $f_\alpha$ is continuous) is each $f_\alpha^{-1}(U_\alpha)$ and thus the inverse image of a basic open set is the unrestricted product of open sets and thus basic open. A similar argument works for the inverse function. $\blacksquare$

June 7, 2010 -

1. Why no love for section 19? (I’m working through the book to study for the math subject GRE.)

Comment by Pat | June 7, 2010 | Reply

2. n/m. Looking at the problems they pretty much seem to be the same. You working out of the first edition?

Comment by Pat | June 7, 2010 | Reply

3. Haha, yeah recheck the title, I just mislabeled it. And no, I am working out of the second edition.

Also, is there really that much top. on the GREs?

Comment by drexel28 | June 7, 2010 | Reply

4. I don’t think there’s too much, which is why I only plan to work my way through the first three chapters. Afterwards will be a crash course in algebra (Fraleigh’s book which seems good). Keep it up, really cool what you’re doing!

Comment by Pat | June 8, 2010 | Reply

5. Thank you! I appreciate that someone actually read this. You’ll notice that this blog mostly used to be theorems, but I just really like doing problems. So, I figured over the summer I’d “try” to do all the problems in the “trifecta”: Topology-Munkres, Algebra-Dummit and Foote (over Herstein,yes), and Analysis-Rudin (baby Rudin). To be fair, I know all of these subjects fairly well. In fact, I’ve been through most of Munkres and Rudin’s book themselves. That said, it’ll still be hard.

P.S. I would not suggest Fraleigh. That book is kind of dumbed down and if you intend to pursue a graduate degree in mathematics (as the GRE reference suggests) it won’t do.

Comment by drexel28 | June 8, 2010 | Reply