# Abstract Nonsense

## Munkres Chapter 2 Section 18

1.

Problem: Show that the normal $\varepsilon-\delta$ formulation of continuity is equivalent to the open set version.

Proof: Suppose that $\left(\mathcal{M},d\right),\left(\mathcal{N},d'\right)$ are metric spaces and for every $\varepsilon>0$ and every $f(x)\in\mathcal{N}$ there exists some $\delta>0$ such that $f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))$. Then, given an open set $U\subseteq \mathcal{N}$ we have that $f^{-1}(U)$. To see this let $x\in f^{-1}(U)$ then $f(x)\in U$ and since $U$ is open by hypothesis there exists some open ball $B_\varepsilon(f(x))$ such that $B_{\varepsilon}(f(x))\subseteq U$ and thus by assumption of $\varepsilon-\delta$ continuity there is some $\delta>0$ such that $\displaystyle f\left(B_{\delta}(x)\right)\subseteq B_{\varepsilon}(f(x))\subseteq U$ and so $B_{\delta}\subseteq f^{-1}(U)$ and thus $x$ is an interior point of $f^{-1}(U)$.

Conversely, suppose that the preimage of an open set is always open and let $f(x)\in\mathcal{N}$ and $\varepsilon>0$ be given. Clearly $B_{\varepsilon}(f(x))$ is open and thus $f^{-1}\left(B_{\varepsilon}(f(x))\right)$ is open. So, since $x\in f^{-1}\left(B_{\varepsilon}(f(x))\right)$ there exists some $\delta>0$ such that $B_{\delta}(x)\subseteq f^{-1}\left(B_{\varepsilon}(f(x))\right)$ and so

$f\left(B_{\delta}(x)\right)\subseteq f\left(f^{-1}\left(B_{\varepsilon}(f(x))\right)\right)\subseteq B_{\varepsilon}(f(x))$

$\blacksquare$

2.

Problem: Suppose that $f:X\to Y$ 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 $f(A)$?

Proof: No. Consider $(-1,0)\cup(0,1)$ with the suspace topology inherited from $\mathbb{R}$ with the usual topology. Define

$f:(-1,0)\cup(0,1)\to D:x\mapsto\begin{cases}0\quad\text{if}\quad x\in(-1,0)\\ 1\quad\text{if}\quad x\in(0,1)\end{cases}$

This is clearly continuous since $f^{-1}(\{1\})=(-1,0)$ and $f^{-1}(\{1\})=(0,1)$ which are obviously open. But, notice that $\frac{-1}{2}$ is a limit point for $(-1,0)$ since given a neighborhood $N$ of $\frac{-1}{2}$ we must have that there is some $(a,b)\cap \left((-1,0)\cup(0,1)\right)\cap \subseteq N$ which contains it. But, $f\left(\frac{-1}{2}\right)=\{0\}$ is not a limit point for $f\left((-1,0)\right)=\{0\}$ since that set has no limit points. $\blacksquare$

3.

Problem: Let $X$ and $X'$ denote a singlet set in the two topologies $\mathfrak{J}$ and $\mathfrak{J}'$ respectively. Let $\text{id}:X'\to X$ be the identity function. Show that

a) $\text{id}$ is continuous if and only if $\mathfrak{J}'$ is finer than $\mathfrak{J}$

b) $\text{id}$ is a homeomorphism if and only if $\mathfrak{J}=\mathfrak{J}'$

Proof:

a) Assume that $\text{id}$ is continuous then given $U\in\mathfrak{J}$ we have that $\text{id}^{-1}(U)=U\in\mathfrak{J}'$. Conversely, if $\mathfrak{J}'$ is finer than $\mathfrak{J}$ we have that given $U\in\mathfrak{J}$ that $\text{id}^{-1}(U)=U\in\mathfrak{J}'$

b) If $\text{id}$ is a homeomorphism we see that both it and $\text{id}^{-1}=\text{id}:X\to X'$ are continuous and so mimicking the last argument we see that $\mathfrak{J}\subseteq\mathfrak{J}'$ and $\mathfrak{J}'\subseteq\mathfrak{J}$. Conversely, if $\mathfrak{J}=\mathfrak{J}'$ then we now that

$U\in\mathfrak{J}\text{ iff }U\in\mathfrak{J}'$ or equivalently that $U\text{ is open in }X\text{ iff }\text{id}(U)=U\text{ is open in }X'$

which defines the homeomorphic property. $\blacksquare$

4.

Problem: Given $x_0\in X$ and $y_0\in Y$ show that the maps $f:X\to X\times Y$ and $g:X\times Y\to Y$ given by $f:x\mapsto (x,y_0)$ and $g:y\mapsto (x_0,y)$ are topological embeddings.

Proof: Clearly $f$ and $g$ are continuous since the projection functions are the identity and constant functions. They are clearly injective for if, for example, $f(x)=(x,y_0)=(x',y_0)=f(x')$ then by definition of an ordered pair we must have that $x=x'$.  Lastly, the inverse function is continuous since $f^{-1}:X\times \{y_0\}\to X:(x,y_0)\mapsto x$ is the restriction of the projection to $X\times\{y_0\}$. The same is true for $g$. $\blacksquare$

5.

Problem: Show that with the usual subspace topology $[0,1]\approx[a,b]$ and $(0,1)\approx(a,b)$.

Proof: Define $f:[0,1]\to[a,b]:x\mapsto (b-a)+a$ and $g:(0,1)\to(a,b):x\mapsto (b-a)+a$. These are easily both proven to be homeomorphisms. $\blacksquare$

6.

Problem: Find a function $f:\mathbb{R}\to\mathbb{R}$ which is continuous at precisely one point.

Proof: Define

$f:\mathbb{R}\to\mathbb{R}:\begin{cases}x\quad\text{if}\quad x\in\mathbb{Q}\\ 0\quad\text{if}\quad x\notin\mathbb{Q}\end{cases}$

Suppose that $f$ is continuous at $x_0$, then choosing sequences $\{q_n\}_{n\in\mathbb{N}},\{i_n\}_{n\in\mathbb{N}}$ of rational and irrationals numbers respectively both converging to $x_0$. We see by the limit formulation of metric space continuity that

$x_0=\lim\text{ }q_n=\lim\text{ }f(q_n)=f(x_0)=\lim\text{ }f(i_n)=\lim\text{ }0=0$

And so if $f$ were to be continuous anywhere it would have to be at $0$. To show that it is in fact continuous at $0$ we let $\varepsilon>0$ be given then choosing $\delta=\varepsilon$ we see that $|x|<\delta\implies |f(x)|\leqslant |x|<\delta=\varepsilon$ from where the conclusion follows since this implies that $\displaystyle \lim_{x\to 0}f(x)=0=f(0)$. $\blacksquare$

7.

Problem:

a) Suppose that $f:\mathbb{R}\to\mathbb{R}$ is “continuous from the right”, that is, $\displaystyle \lim_{x\to a^+}f(x)=f(a)$ for each $a\in\mathbb{R}$. Show that $f$ is continuous when considered as a function from $\mathbb{R}_\ell$ to $\mathbb{R}$.

b) Can you conjecture what kind of functions $f:\mathbb{R}\to\mathbb{R}$ are continuous when considered as maps as $\mathbb{R}\to\mathbb{R}_\ell$. As maps from $\mathbb{R}_\ell$ to $\mathbb{R}_\ell$?

Proof:

a) Note that by the assumption that $\displaystyle \lim_{x\to a^+}f(x)=f(a)$ we know that for every $\varepsilon>0$ there exists some $\delta>0$ such that $0\leqslant x-a<\delta$ implies that $|f(x)-f(a)|<\varepsilon$. So, let $U\subseteq\mathbb{R}$ be open and let $a\in f^{-1}(U)$. Then, $f(a)\in U$ and since $U$ is open we see that there is some $\varepsilon>0$ such that $B_{\varepsilon}(f(a))\subseteq U$. But, by assumption there exists some $\delta>0$ such that $0\leqslant x-a<\delta\implies f(x)\in B_{\varepsilon}(f(a))$. But, $\left\{x: 0\leqslant x-a<\delta\right\}=[a,a+\delta)$ and thus $f\left([a,a+\delta)\right)\subseteq B_{\varepsilon}(f(a))\subseteq U$ and thus $[a,a+\delta)\subseteq f^{-1}(U)$ and so $a$ is an interior point for $f^{-1}(U)$ from where it follows that $f^{-1}(U)$ is open and thus $f$ is continuous.

b) I’m not too sure, and not too concerned right now. My initial impression is that if $f:\mathbb{R}\to\mathbb{R}_\ell$ is continuous then $f^{-1}([a,b))$ is open which should be hard to do. Etc.

8.

Problem: Let $Y$ be an ordered set in the order topology. Let $f,g:X\to Y$ be continuous.

a) Show that the set $\Omega=\left\{x\in X:f(x)\leqslant g(x)\right\}$ is closed in $X$

b) Let $h:X\to Y:x\mapsto \max\{f(x),g(x)\}$. Show that $h$ is continuous.

Proof:

a) Let $x_0\notin\Omega$ then $f(x_0)>g(x_0)$. Suppose first that there is no $g(x_0)<\xi and consider

$f^{-1}\left(g(x_0),\infty)\right)\cap g^{-1}\left((-\infty,f(x_0)\right)=U$

This is clearly open in $X$ by the continuity of $f,g$ and $x_0$ is contained in it. Now, to show that $U\cap \Omega=\varnothing$ let $z\in U$ then $f(z)\in f\left(f^{-1}\left(g(x_0),\infty)\right)\cap g^{-1}\left(-\infty,f(x_0)\right)\right)$ which with simplification gives the important part that $f(z)\in (g(x_0),\infty)$ and so $f(z)>g(x_0)$ but since there is no $\xi$ such that $g(x_0)<\xi this implies that $f(z)\geqslant f(x_0)$. Similar analysis shows that $g(z)\in (-\infty,f(x_0))$ and since there is no $\xi$ as was mentioned above this implies that $g(z)\leqslant g(x_0)$. Thus, $g(z)\leqslant g(x_0) and thus $z\notin\Omega$.

Now, suppose that there is some $\xi$ such that $g(x_0)<\xi then letting $V=f^{-1}(\xi,\infty)\cap g^{-1}(-\infty,\xi)$ we once again see that $V$ is open and $x_0\in V$. Furthermore, a quick check shows that if $z\in V$ that $f(z)\in(\xi,\infty)$ and so $f(z)>\xi$ and $g(z)\in(-\infty,\xi)$ and so $g(z)<\xi$ and so $f(z)>g(z)$ so that $z\notin\Omega$. The conclusion follows

b) Let $\Omega_f=\left\{x\in X:f(x)\geqslant g(x)\right\}$ and $\Omega_g=\left\{x\in X:g(x)\geqslant f(x)\right\}$. As was shown in a) both $\Omega_f,\Omega_g$ are closed and thus define

$f\sqcup g:X=\left(\Omega_f\cup\Omega_g\right)\to Y:x\mapsto\begin{cases}f(x)\quad\text{if}\quad x\in\Omega_f\\ g(x)\quad\text{if}\quad x\in\Omega_g\end{cases}$

Notice that since $f,g$ are both assumed continuous and $f\mid_{\Omega_g\cap\Omega_f}=g\mid_{\Omega_f\cap\Omega_g}$ that we may conclude by the gluing lemma that $f\sqcup g$ is in fact continuous. But, it is fairly easy to see that $f\sqcup g=\max\{f(x),g(x)\}$ $\blacksquare$

9.

Problem: Let $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ be a collection of subset of $X$; let $\displaystyle X=\bigcup_{\alpha\in\mathcal{A}}U_\alpha$. Let $f:X\to Y$ and suppose that $f\mid_{U_\alpha}$ is continuous for each $\alpha\in\mathcal{A}$

a) Show that if the collection $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is finite each set $U_\alpha$ is closed, then $f$ is continuous.

b) Find an example where the collection $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is countable and each $U_\alpha$ is closed but $f$ is not continuous.

c) An indexed family of sets $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is said to be locally finite if each point of $X$ has a neighborhood that intersects only finitely many elements of $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$. Show that if the family $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is locally finite and each $U_\alpha$ is closed then $f$ is continuous.$Proof: a) This follows since if $V\subseteq Y$ is closed then it is relatively easy to check that $\displaystyle f^{-1}(V)=\bigcup_{\alpha\in\mathcal{A}}\left(f\mid_{U_\alpha}\right)^{-1}(V)$ but since each $f\mid_{U_\alpha}$ is continuous we see that $\left(f\mid_{U_\alpha}\right)^{-1}(V)$ is closed in $U_\alpha$. But, since each $U_\alpha$ is closed in $X$ it follows that each $\left(f\mid_{U_\alpha}\right)^{-1}(V)$ is closed in $X$. Thus, $f^{-1}(V)$ is the finite union of closed sets in $X$, and thus closed. b) Give $[0,1]$ the subspace topology inherited from $\mathbb{R}$ with the usual topology and consider $\left\{f_n\right\}_{n\in\mathbb{N}-\{1,2\}}$ with $f_n=\iota_{[0,1-\frac{1}{n}]}:\left[0,1-\tfrac{1}{n}\right]\to[0,1]:x\mapsto x$ Clearly each $f_n$ i Lemma: Let $Y$ be any topological space and $\left\{V_\beta\right\}_{\beta\in\mathcal{B}}$ be a locally finite collection of subsets of $Y$. Then, $\displaystyle \bigcup_{\beta\in\mathcal{B}}\overline{V_\beta}=\overline{\bigcup_{\beta\in\mathcal{B}}V_\beta}$ Proof: The left hand inclusion is standard, so it suffices to prove the right inclusion. So, let $\displaystyle x\in\overline{\bigcup_{\beta\in\mathcal{B}}V_\beta}$ since the collection of sets is locally finite there exists some neighborhood $N$ of $x$ such that it intersects only finitely many, say $V_{\beta_1},\cdots,V_{\beta_n}$, elements of the collection. So, suppose that $x\notin \left(\overline{V_{\beta_1}}\cup\cdots\cup \overline{V_{\beta_n}}\right)$ then $N\cap-\left(\overline{V_{\beta_1}}\cup\cdots\cup\overline{V_{\beta_n}}\right)$ is a neighborhood of $x$ which does not intersect $\displaystyle \bigcup_{\beta\in\mathcal{B}}V_\beta$ contradicting the assumption it is in the closure of that set. $\blacksquare$ Now, once again we let $V\subseteq Y$ be closed and note that $\displaystyle f^{-1}(V)=\bigcup_{\alpha\in\mathcal{A}}\left(f\mid_{U_\alpha}\right)^{-1}(V)$ and each $\left(f\mid_{U_\alpha}\right)^{-1}(V)$ is closed in $U_\alpha$ and since $U_\alpha$ is closed in $X$ we see that $\left(f\mid_{U_\alpha}\right)^{-1}(V)$ is closed in $X$. So, noting that $\left(f\mid_{U_\alpha}\right)^{-1}(V)\subseteq U_\alpha$ it is evident from the assumption that $\left\{U_\alpha\right\}_{\alpha\in\mathcal{A}}$ is locally finite in $X$ that so is $\left\{\left(f\mid_{U_\alpha}\right)^{-1}(V)\right\}_{\alpha\in\mathcal{A}}$ and thus (for notational convenience) letting $F_\alpha=\left(f\mid_{U_\alpha}\right)^{-1}(V)$ the above lemma implies that $\displaystyle \overline{f^{-1}(V)}=\overline{\bigcup_{\alpha\in\mathcal{A}}F_\alpha}=\bigcup_{\alpha\in\mathcal{A}}\overline{F_\alpha}=\bigcup_{\alpha\in\mathcal{A}}F_\alpha=f^{-1}(V)$ From where it follows that the preimage of a closed set under $f$ is closed. The conclusion follows. $\blacksquare$ 10. Problem: Let $f:A\to B$ and $g:C\to D$ be continuous functions. Let us define a map $f\times g:A\times C\to B\times D$ by the equation $(f\times g)(a\times c)=f(a)\times g(c)$. Show that $f\times g$ is continuous. Proof: This follows from noting the two projections of $f\times g$ are $\pi_1\circ(f\times g):A\times B\to C:a\times b\mapsto f(a)$ and $\pi_2\circ(f\times g):A\times B\to D:a\times b\mapsto f(b)$. But, both of these are continuous since $\left(\pi_1(f\times g)\right)^{-1}(U)=f^{-1}(U)\times B$. To see this we note that $x\in f^{-1}(U)\times B$ if and only if $x\in f^{-1}(U)$ which is true if and only if $f(x)=\left(\pi_1\circ(f\times g)\right)(x)\in U$ or in other words $x\in \left(\pi_1\circ(f\times g)\right)^{-1}(U)$. Using this we note that the preimage an open set in $C$ will be the product of open sets by the continuity of $f$. It clearly follows both projections, and thus the function itself are continuous. $\blacksquare$ 11. Problem: Let $F:X\times Y\to Z$. We say that $F$ is continuous in eahc variable separately if for each $y_0\in Y$, the map $h:X\to Z:x\mapsto F(x\times y_0($ is continuous and for each $x_0\in X$ the map $j:Y\to Z:y\mapsto F(x_0\times y)$ is continuous. Show that if $F$ is continuous then $F$ is continuous in each variable separately. Proof: If $F$ is continuous then clearly it is continuous in each variable since if we denote by $G_{y_0}$ the mapping $G_{y_0}:X\to Z:x\mapsto F(x\times y_0)$ we see that $G_{y_0}=H_{y_0}\circ(F\mid_{X\times\{y_0\}})$ where $H_{y_0}:X\to X\times Y:x\mapsto x\times y_0$ but the RHS is the composition of continuous maps and thus continuous. A similar analysis holds for the other variable. 12. Problem: Let $F:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ be given by $\displaystyle F(x\times y)=\begin{cases} \frac{xy}{x^2+y^2}&\mbox{if}\quad x\times y\ne 0\times 0\\ 0 &\mbox{if} \quad x\times y=0\times0\end{cases}$ a) Show that $F$ is continuous in each variable separately. b) Compute $g:\mathbb{R}\to\mathbb{R}:x\mapsto F(x\times x)$. c) Show that $F$ is not continuous Proof: a) Clearly both $F(x\times y_0)$ and $F(x_0\times y)$ are continuous for $x,y\ne 0$ since they are the quotient of polynomials and the denominator is non-zero. Lastly, they are both continuous at $x,y=0$ since it is trivial to check that$

$\displaystyle 0=F(0\times y_0)=F(x_0\times 0)=\lim_{x\to 0}F(x\times y_0)=\lim_{y\to 0}F(x_0\times y)$

b) Evidently

$\displaystyle g(x)=F(x\times x)=\begin{cases}\frac{1}{2}\quad\text{if}\quad x\times x\ne 0\\ 0\quad\text{if}\quad x\times x=0\end{cases}$

c) This clearly proves that $F(x\times y)$ is not continuous with $\mathbb{R}^2$ is not continuous since if $\Delta$ is the diangonal we have that

$\displaystyle \lim_{(x,y)\in\Delta\to (0,0)}F(x\times y)=\frac{1}{2}\ne F(0\times 0)$

and so in particular

$\displaystyle \lim _{(x,y)\to(0,0)}F(x\times y)\ne F(0\times 0)$

13.

Problem: Let $A\subseteq X$; let $f:A\to X$ be continuous and let $Y$ be Hausdorff. Prove that if $f$ may be extended to a continuous function $\overset{\sim}{f}:\overline{A}\to Y$, then $\overset{\sim}{f}$ is uniquely determined by $f$.

Proof: I am not too sure what this question is asking, but assuming it’s asking that if this extension existed it’s unique we can do this two ways

Way 1(fun way!):

Lemma: Let $X$ be any topological space and $Y$ a Hausdorff space. Suppose that $\varphi,\psi:X\to Y$ are continuous and define $A(\varphi,\psi)=\left\{x\in X:\varphi(x)=\psi(x)\right\}$. Then, $A(\varphi,\psi)$ is closed in $X$

Proof: Note that $\varphi\oplus\psi:X\to Y\times Y:x\mapsto (\varphi(x),\psi(x))$ is clearly continuous since $\pi_1\circ(\varphi\oplus\psi)=\varphi$ and $\pi_2\circ(\varphi\oplus\psi)=\psi$. It is trivial then to check that $\displaystyle A(\varphi,\psi)=\left(\varphi\oplus\psi\right)^{-1}(\Delta_Y)$ and since $Y$ is Hausdorff we have that $\Delta_Y\subseteq Y\times Y$ is closed and the conclusion follows. $\blacksquare$

From this we note that if $\varphi,\psi$ agree on $D\subseteq X$ such that $\overline{D}=X$ we have that

$X\supseteq A(\varphi,\psi)=\overline{A(\varphi,\psi)}\supseteq\overline{D}=X$

From where it follows that $A(\varphi,\psi)=X$ and so $\varphi=\psi$. So, thinking of $\overline{A}$ as a subspace of $X$ we see that $\text{cl}_{\overline{A}}\text{ }A=Y\cap\text{cl}_{X}\text{ }A=\overline{A}$ and thus clearly $A$ is dense in $\overline{A}$. So, the conclusion readily follows by noting that if $\overset{\sim}{f_1},\overset{\sim}{f_2}$ are two continuous extensions then by definition $A\left(\overset{\sim}{f_1},\overset{\sim}{f_2}\right)\supseteq A$.

Way 2(unfun way): Let $\overset{\sim}{f_1},\overset{\sim}{f_2}$ be two extensions of $f$ and suppose there is some $x\in\overline{A}-A(\varphi,\psi)$. Clearly $x\notin A$ and thus $x$ is a limit point of $A$. So, by assumption $\overset{\sim}{f_1}(x)\ne\overset{\sim}{f_2}(x)$ and so using the Hausdorffness of $Y$ we may find disjoint neighborhoods $U,V$ of them respectively. Thus, $\overset{\sim}{f_1}^{-1}(U),\overset{\sim}{f_2}^{-1}(V)$ are neighborhoods of $x$ in $X$. Thus, $\overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V)$ is a neighborhood of $x$. But, clearly there can be no $y\in A\cap\left(\overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V)\right)$ otherwise $\overset{\sim}{f_1}(y)=\overset{\sim}{f_2}(y)\in U\cap V$. It follows that $\overset{\sim}{f_1}^{-1}(U)\cap\overset{\sim}{f_2}^{-1}(V)$ is a neighborhood of $x$ disjoint from $A$ which contradicts the density of $A$ in $\overline{A}$.  The conclusion follow. $\blacksquare$

May 28, 2010 -

1. A comment on your solution to #2: it appears that defining X=(-1,0)\cup(0,1) is unnecessary, since X=(-1,0) would work just as well. I suspect that your example is a holdover from an early draft of your solution when you were likely trying to consider 0 as the limit point.

At any rate, you also have two typos: f^{-1}(\{0\})=(-1,0) as opposed to f^{-1}\{(1\})=(-1,0), and you have an extra \cap in your third to last tex string.

Comment by soiteroo | July 3, 2011 | Reply

• Ah, thank you! You are one-hundred percent correct! It was a holdover from a previous draft. I had done all of these far before I posted them up here and was anxious to get them down, they are also smattered with problems from other texts by accident. I will fix that typo then, thanks!

Comment by Alex Youcis | July 8, 2011 | Reply

2. Also, though it’s essentially what you’re doing I think 9a is even more trivial when you recognize it as induction on the number of pasted-together sets in the pasting lemma.

[I’m not trying to be disagreeable; I’m going through the same project as you and I’ve been comparing my solutions to yours.]

Comment by soiteroo | July 3, 2011 | Reply

• Friend,

Thanks for the input! I see what you’re saying, and I agree that induction would work fine! I think it is more of a stylistic difference than anything.

I don’t find you disagreeable in the slightest, I really enjoy when people point out flaws in my solutions/alternate (better) solutions! Math is a communal subject, and any feedback (good or bad) helps me learn.

I wish you luck in your goal of finishing Munkres. I eventually got restless and moved on to other subjects, but I got fairly far (I just didn’t post them). If you have any specific questions about a problem feel free to e-mail me!

Comment by Alex Youcis | July 8, 2011 | Reply