Abstract Nonsense

Crushing one theorem at a time

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<b,c<d,\text{ }a,b,c,d\in\mathbb{Q}\right\} is an open base for \mathbb{R}^2

Proof: This follows from the fact that \left\{(a,b):a<b,\text{ }a,b\in\mathbb{Q}\right\} 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<c<b we have that a^2<c^2<b^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<b,\text{ }a,b\in\mathbb{R}\right\} and \left\{(a,b):a<b,\text{ }a,b\in\mathbb{R}\right\} are open bases for \mathbb{R}_\ell and \mathbb{R} we know that \mathfrak{B}=\left\{(a,b)\times[c,d):a<b,c<d,\text{ }a,b,c,d\in\mathbb{R}\right\} 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<d,\text{ }c,d\in\mathbb{R}\right\}\cup\{\varnothing\}.  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<d\text{ }c,d\in\mathbb{R}\right\}\cup\{\varnothing\}

Conversely, let U\in\left\{\{x_0\}\times[c,d):c<d,\text{ }c,d\in\mathbb{R}\right\}\cup\{\varnothing\}. 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<b,\text{ }a,b\in\mathbb{R}\right\}

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}<y<b 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<b. 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}<y_0 and thus (a,b)<(a,y_0) and since a<c 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<x_0<c 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<a\implies (a,b)<(x_0,y) and x_0<c\implies (x_0,y)<(c,d). 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<b,c<d\text{ }a,b,c,d\in I\right\} 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

Advertisements

May 21, 2010 - Posted by | Fun Problems, Topology | , , , , ,

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: