# Abstract Nonsense

## Munkres Chapter 2 Section 2

1.

Problem: Show that if $Y$ is a subspace of $X$, and $A$ is a subspace of $Y$, the the topology it inherits as a subspace of $Y$ is the same as it inherits as a subspace of $X$.

Proof: Let $\mathfrak{J}_X$ and $\mathfrak{J}_Y$ be the induced topologies on $A$ as a subspace of $X,Y$ respectively. Let $U\in\mathfrak{J}_X$ then $U=A\cap V$ for some open set $V$ in $X$. But, $A\cap V=Y\cap A\cap V=A\cap (V\cap Y)$ and since $V\cap Y$ is open in $Y$ it follows that $U\in\mathfrak{J}_Y$. Conversely, suppose that $U\in\mathfrak{J}_Y$ then $U=A\cap V$ for some open set $V$ in $Y$. But, since $Y$ is a subspace of $X$ we know that $V=W\cap Y$ for some open set $W$ in $X$. Thus, $U=A\cap V=A\cap W\cap Y=A\cap W$ and thus $A\in\mathfrak{J}_X$. The conclusion follows. $\blacksquare$

2.

Problem: If $\mathfrak{J},\mathfrak{J}'$ are topologies on $X$ and $\mathfrak{J}'$ is strictly finer than $\mathfrak{J}$, what can you say about the corresponding subspace topologies on the subset $Y$ of $X$?

Proof: Let $\mathfrak{I}$ and $\mathfrak{I}'$ be the subspace topologies on $Y$ inherited from $\mathfrak{J},\mathfrak{J}'$ respectively. Let $U\in\mathfrak{I}$ then $U=Y\cap V$ for some $V\in\mathfrak{J}\subseteq\mathfrak{J}'$ and thus $U\in\mathfrak{I}'$. Thus, $\mathfrak{I}\subseteq\mathfrak{I}'$. The inclusion need not be strict. For example consider the set $X=\{a,b,c\}$ with the two topologies

$\mathfrak{J}=\left\{\varnothing,X\right\}$

$\mathfrak{J}'=\left\{\varnothing,\{a\},X\right\}$

Now, consider $Y=\{b\}\subseteq X$ then

$\mathfrak{I}=\left\{\varnothing\cap\{b\},X\cap\{b\}\right\}=\left\{\varnothing,Y\right\}$

$\mathfrak{I}'=\left\{\varnothing\cap\{b\},\{a\}\cap\{b\},X\cap\{b\}\right\}=\left\{\varnothing,Y\right\}$

And thus $\mathfrak{I}=\mathfrak{I}'$. The inclusion will be strict if $Y$ intersects every element of $\mathfrak{J}$. $\blacksquare$

3.

Problem: Consider the set $Y=[-1,1]$ as a subspace of $\mathbb{R}$. Which of the following sets are open in $Y$?

$A=\left\{x:\frac{1}{2}<|x|<1\right\}$

$B=\left\{\frac{1}{2}<|x|\leq1\right\}$

$C=\left\{x:\frac{1}{2}\leq|x|\leq1\right\}$

$D=\left\{x:\frac{1}{2}\leq|x|\leq1\right\}$

$E=\left\{x:0<|x|<1\text{ and }\frac{1}{x}\notin\mathbb{N}\right\}$

Proof:

We only bother to state which are open in $Y$ since the results should be clear for $\mathbb{R}$ to any student in pre-algebra.

$A$ is open sine $A=Y\cap \left((-1,\frac{-1}{2})\cup(\frac{1}{2},1)\right)$.

$B$ is open since $B=Y\cap\left( (-\infty,\frac{-1}{2})\cup(\frac{1}{2},\infty)\right)$

$C$ is not open. To see this suppose that it was then $Y-C=\left(\frac{-1}{2},\frac{1}{2}\right)\cup\{-1,1\}$ is closed in $Y$, but since $Y$ is closed in $\mathbb{R}$ this would imply that $(\frac{-1}{2},\frac{1}{2})\cup\{-1,1\}$ was closed in $\mathbb{R}$.

$D$ is not open since if it were it’s complement $Y-D=(\frac{-1}{2},\frac{1}{2})$ would be closed in $Y$ and since $Y$ is closed in $X$ it follows $(\frac{-1}{2},\frac{1}{2})$ is closed in $\mathbb{R}$.

$E$: This is open since $E=Y\cap (-1,1)\cap\left(\mathbb{R}-K\right)$ the last of which is open. $\blacksquare$

4.

Problem: A map $f:X\to Y$ is said to be an open map if for every open set $U$ of $X$, the set $f(U)$ is open in $Y$. Show that $\pi_1:X\times Y\to X$ and $\pi_2:X\times Y\to Y$ are open maps.

Proof: Assuming that $X\times Y$ has the product topology we know that given an open set $W\subseteq X\times Y$ we know that $\displaystyle W=\bigcup_{\alpha\in\mathcal{A}}\left(U_\alpha\times V_\alpha\right)$ where $U_\alpha,V_\alpha$ are open in $X,Y$ respectively. Thus,

$\displaystyle \pi_1\left(W\right)=\pi_1\left(\bigcup_{\alpha\in\mathcal{A}}\left(U_\alpha\times V_\alpha\right)\right)=\bigcup_{\alpha\in\mathcal{A}}\pi_1\left(U_\alpha\times V_\alpha\right)=\bigcup_{\alpha\in\mathcal{A}}U_\alpha$

$\displaystyle \pi_2\left(W\right)=\pi_2\left(\bigcup_{\alpha\in\mathcal{A}}\left(U_\alpha\times V_\alpha\right)\right)=\bigcup_{\alpha\in\mathcal{A}}\pi_2\left(U_\alpha\times V_\alpha\right)=\bigcup_{\alpha\in\mathcal{A}}V_\alpha$

Which are open in $X,Y$ respectively. $\blacksquare$

5.

Problem: Let $X$ and $X'$ denote a simle set in the topologies $\mathfrak{J},\mathfrak{J}'$ respectively; let $Y$ and $Y'$ denote a singe set in the topologies $\mathfrak{U},\mathfrak{U}'$ respectively. Assume that these sets are non-empty.  Prove that if $\mathfrak{J}'\supseteq\mathfrak{J}$ and $\mathfrak{U}'\supseteq\mathfrak{U}$ then the product topology $X'\times Y'$ is finer than the product topology on $X\times Y$

Proof:It suffices to check that each basic element in $X'\times Y'$ has a basic element in $X\times Y$ contained within it which contains and arbitrary point. So, let $U'\times V'$ be open in $X'\times Y'$ and let $(x,y)\in U'\times V'$\$. Then, we see by the previous problem that $x\in U'$ and $y\in V'$ and thus since the topologies on $X,Y$ are finer than those on $X',Y'$ there are basic open sets $U,V$ in $X,Y$ respectively such that $x\in U\subseteq U'$ and $y\in V\subseteq V'$. Thus, $(x,y)\subseteq U\times V\subseteq U'\times V'$ and since $U\times V$ is basic open in $X\times Y$ the conclusion follows. $\blacksquare$

Problem: Show that $\left\{(a,b)\times(c,d):a is an open base for $\mathbb{R}^2$

Proof: This follows from the fact that $\left\{(a,b):a is an open base for $\mathbb{R}$ and $\mathbb{R}^2$ has the product topology. $\blacksquare$

7.

Problem: Let $X$ be an ordered set. If $Y$ is a proper subset of $X$ that is convex in $X$ does it follow that $Y$ is an interval or a ray in $X$?

Proof: No. Consider $\mathbb{Q}$ with the usual ordering and the set $\left\{q\in\mathbb{Q}:q^2 <2\right\}=U$. This is clearly convex since if $a,b\in U$ then for any $a we have that $a^2 and thus $c\in U$ or in other words $(a,b)\subseteq U$. But, it is not an interval or a ray. Clearly if it were either it must be a ray, so assume that $U=(-\infty,\alpha)$ or $U=(-\infty,\beta]$. In both cases we must have that $\alpha,\beta<\sqrt{2}$ and thus by the density of the rationals in $\mathbb{R}$ we may find $\gamma$ such that $|\alpha|,|\beta|<\gamma<\sqrt{2}$ or in other words that $\alpha^2,\beta^2<\gamma^2<2$ and thus $\gamma\notin(-\infty,\alpha),(-\infty,\beta]$. $\blacksquare$

8.

Problem: If $L$ is a straight line in the plane describe the topology $L$ inherets as a subspace of $\mathbb{R}\times\mathbb{R}_\ell$ and as a subspace of $\mathbb{R}_\ell\times\mathbb{R}_\ell$. In each case it is a familiar topology.

Proof: Clearly as was shown since $\left\{[a,b):a and $\left\{(a,b):a are open bases for $\mathbb{R}_\ell$ and $\mathbb{R}$ we know that $\mathfrak{B}=\left\{(a,b)\times[c,d):a is an open base for $\mathbb{R}\times\mathbb{R}_\ell$. Now, if you’ll recall the subspace topology on $L$ can be described by $\mathfrak{B}'=\left\{L\cap B:B\in\mathfrak{B}\right\}$. So, we break this into three cases based on what kind of line $L$ is.

So, firstly suppose that $L$ is a vertical line then $L=\{x_0\}\times\mathbb{R}$ for some $x_0\in\mathbb{R}$. So, we now claim $\mathfrak{B}'=\left\{\{x_0\}\times[c,d):c.  So, let $U=L\cap \left((a,b)\times[c,d)\right)\in\mathfrak{B}'$ then we have that

$U=L\cap\left((a,b)\times[c,d)\right)=\left(\{x_0\}\times\mathbb{R}\right)\cap\left((a,b)\times[c,d)\right)=\left(\{x_0\}\cap(a,b)\right)\times [c,d)$

Which, equals

$\displaystyle \begin{cases}\varnothing\quad\text{if}\quad x_0\notin(a,b)\\ \{x_0\}\times[c,d)\quad\text{if}\quad x_0\in(a,b)\end{cases}$

Either way $U\in\left\{\{x_0\}\times[c,d):c

Conversely, let $U\in\left\{\{x_0\}\times[c,d):c. Then,

$\displaystyle \begin{cases} U=L\cap\left((x_0-2,x_0-1)\times[0,1)\right)\quad\text{if}\quad U=\varnothing\\ U=L\cap\left((x_0-1,x_0+1)\times[c,d)\right)\quad\text{if}\quad U=\{x_0\}\times[c,d)\end{cases}$

Either way we see that $U\in\mathfrak{B}'$.

A similar analysis shows that if $L=\mathbb{R}\times\{y_0\}$ that $\mathfrak{B}'=\left\{(a,b)\times\{y_0\}:a

We now break this into the last case where $L=\left\{(x,\alpha x):x\in\mathbb{R}\right\}$ for some $\alpha\in\mathbb{R}-\{0\}$. So, once again we know that the subspace topology on $L$ will be that generated by $\mathfrak{B}'=\left\{B\cap L:B\in\mathfrak{B}\right\}$. We now get slightly more informal. The key point is that since none of the corners of the rectangle $(a,b)\times[c,d)$ are included the line, if it intersects, it intersects an interval like region of the line. Thus, a quick breakdown into the cases where it intersects the solid line $(a,b)\times\{c\}$ and when it doesn’t should quickly convince you that the topology on $L$ will end up being the lower-limit topology.

Now, for the case with $\mathbb{R}_\ell\times\mathbb{R}_\ell$ we have quite a different scenario. For example consider the anti-diagonal $-\Delta=\left\{(x,-x):x\in\mathbb{R}\right\}$ and consider for each fixed $x_0\in -\Delta$ the basic open set $[x_0,x_0+1)\times[x_0,x_0+1)$ clearly then $-\Delta\cap [x_0,x_0+1)\times[x_0,x_0+1)=\{x_0\}$ from where it follows that $-\Delta$ is in fact a discrete space. In fact for any line that is non-horizontal and non-vertical we may find for each point on the line a basic open set whose intersection with the line is that single point. In other words the lines in $\mathbb{R}_\ell\times\mathbb{R}_\ell$ inherit the discrete topology as subspaces. For horizontal and vertical lines they still have the lower limit topology. $\blacksquare$

9.

Problem: Show that the dictionary order topology on the set $\mathbb{R}\times\mathbb{R}$ is the same as the product topology on $\mathbb{R}_d\times\mathbb{R}$ where $\mathbb{R}_d$ denotes $\mathbb{R}$ with the discrete topology. Compare this topology to the usual topology on $\mathbb{R}^2$

Proof: Let $(x_0,y_0)\in \mathbb{R}_d\times\mathbb{R}$ and let $\{x_0\}\times(a,b)$ be a basic open neighborhood of it. Then, consider the set $\left(\left(x_0,\frac{a+y_0}{2}\right),(x_0,b)\right)$ clearly this is open in $\mathbb{R}\times\mathbb{R}$ with the lexicographic ordering and if $(x,y)\in\left(\left(x_0,\frac{a+y_0}{2}\right),(x_0,b)\right)$ we must have that $\left(x_0,\frac{y_0+a}{2}\right)<(x,y)<(x_0,b)$ which implies that $x=x_0\text{ and }a<\frac{y_0+a}{2} and thus $(x,y)\in\{x_0\}\times(a,b)$. Thus, noting $(x_0,y_0)\in\left(x_0,\frac{y_0+a}{2}\right)$ finishes that portion of the argument. Conversely, let $(x_0,y_0)\in\mathbb{R}\times\mathbb{R}$ and $\left((a,b),(c,d)\right)$ be some basic open neighborhood containing it. If $a=c$ then $\left((a,b),(c,d)\right)=\{a\}\times(b,d)$ and we’re done. So, assume that $a. If $x_0=a$ then consider $\{a\}\times\left(\frac{y_0+b}{2},y_0+1\right)$. Clearly $(a,y_0)\in\{a\}\times\left(\frac{y_0+b}{2},y_0+1\right)$. Now, to see that $\{a\}\times\left(\frac{y_0+b}{2},y_0+1\right)$ we note that $(a,y_0)\in\{a\}\times\left(\frac{y_0+b}{2},y_0+1\right)\implies b<\frac{y_0+b}{2} and thus $(a,b)<(a,y_0)$ and since $a we automatically have that $(a,y)<(c,d)$.

Now, if $x_0=c$ then choosing $\{c\}\times\left(y_0-1,\frac{y_0+d}{2}\right)$ and applying similar techniques works.

Lastly, if $a then choosing $\{x_0\}\times\left(y_0-1,y_0+1\right)$ automatically works since $(x_0,y_0)\in\{x_0\}\times(y_0-1,y_0+1)$ and if $(x_0,y)\in\{x_0\}\times(y_0-1,y_0+1)$ we have that $x_0 and $x_0. Regardless, we’ve found a basic open set in $\mathbb{R}_d\times\mathbb{R}$  which contains $(x_0,y_0)$ and which is contained in $\left((a,b),(c,d)\right)$.

It follows that the each topology is finer than the others, or that the topologies are equal.

The order topology on $\mathbb{R}\times\mathbb{R}$ is finer than that of the usual topology on $\mathbb{R}^2$. Just note that if $(x_0,y_0)\in\mathbb{R}^2$ and $B_\delta((x_0,y_0))$ is an open ball centered at it that $\{x_0\}\times(y_0-\delta,y_0+\delta)$ is a basic open set in $\mathbb{R}\times\mathbb{R}$ containing $(x_0,y_0$ and which is contained inside the ball. The converse is not true. For example consider the origin $(0,0)$ and the basic open neighborhood $\{0\}\times(-1,1)$. Any open ball centered at zero will contain points not on that line from where the conclusion follows. $\blacksquare$

10.

Problem: Let $I=[0,1]$. Compare the product topology on $I\times I$, the dictionary order topology on $I\times I$, and the topology $I\times I$ inherits as a subspace of $\mathbb{R}\times\mathbb{R}$ in the dictionary order topology.

Proof: Let $\mathfrak{J}_1,\mathfrak{J}_2,\mathfrak{J}_3$ denote the topologies in the order they were mentioned. We first claim that $\mathfrak{J}_1\subsetneq\mathfrak{J}_2$. To see this let $(x_0,y_0)\in I\times I$ be arbitrary and note that $\left\{(a,b)\times(c,d):a generates $\mathfrak{J}_1$. Then, if $(x_0,y_0)\in(a,b)\times(c,d)$ we have that $(x_0,y_0)\in\left((x_0,b),(x_0,d)\right)\subseteq(a,b)\times(c,d)$.  To see why the inclusion’s strict we note as in the last example that for example $(\frac{1}{2},\frac{1}{2})\in\left((\frac{1}{2},0),(\frac{1}{2},1)\right)$ and any open square must intersect points of  not in that interval.

Clearly then $\mathfrak{J}_3\subsetneq\mathfrak{J}_2$  as was shown in the book. $\blacksquare$