## Munkres Chapter 2 Section 2

**1.**

**Problem: **Show that if is a subspace of , and is a subspace of , the the topology it inherits as a subspace of is the same as it inherits as a subspace of .

**Proof:** Let and be the induced topologies on as a subspace of respectively. Let then for some open set in . But, and since is open in it follows that . Conversely, suppose that then for some open set in . But, since is a subspace of we know that for some open set in . Thus, and thus . The conclusion follows.

**2.**

**Problem: **If are topologies on and is strictly finer than , what can you say about the corresponding subspace topologies on the subset of ?

**Proof: **Let and be the subspace topologies on inherited from respectively. Let then for some and thus . Thus, . The inclusion need not be strict. For example consider the set with the two topologies

Now, consider then

And thus . The inclusion will be strict if intersects every element of .

**3.**

**Problem:** Consider the set as a subspace of . Which of the following sets are open in ?

**Proof:**

We only bother to state which are open in since the results should be clear for to any student in pre-algebra.

is open sine .

is open since

is not open. To see this suppose that it was then is closed in , but since is closed in this would imply that was closed in .

is not open since if it were it’s complement would be closed in and since is closed in it follows is closed in .

: This is open since the last of which is open.

**4.**

**Problem:** A map is said to be an open map if for every open set of , the set is open in . Show that and are open maps.

**Proof: ****Assuming that has the product topology we know that given an open set we know that where are open in respectively. Thus,**

**Which are open in respectively. **

**5.**

**Problem: **Let and denote a simle set in the topologies respectively; let and denote a singe set in the topologies respectively. Assume that these sets are non-empty. ** **Prove that if and then the product topology is finer than the product topology on

**Proof:**It suffices to check that each basic element in has a basic element in contained within it which contains and arbitrary point. So, let be open in and let $. Then, we see by the previous problem that and and thus since the topologies on are finer than those on there are basic open sets in respectively such that and . Thus, and since is basic open in the conclusion follows.

**Problem: **Show that is an open base for

**Proof:** This follows from the fact that is an open base for and has the product topology.

**7.**

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

**Proof: **No. Consider with the usual ordering and the set . This is clearly convex since if then for any we have that and thus or in other words . But, it is not an interval or a ray. Clearly if it were either it must be a ray, so assume that or . In both cases we must have that and thus by the density of the rationals in we may find such that or in other words that and thus .

**8.**

**Problem: **If is a straight line in the plane describe the topology inherets as a subspace of and as a subspace of . In each case it is a familiar topology.

**Proof: **Clearly as was shown since and are open bases for and we know that is an open base for . Now, if you’ll recall the subspace topology on can be described by . So, we break this into three cases based on what kind of line is.

So, firstly suppose that is a vertical line then for some . So, we now claim . So, let then we have that

Which, equals

Either way

Conversely, let . Then,

Either way we see that .

A similar analysis shows that if that

We now break this into the last case where for some . So, once again we know that the subspace topology on will be that generated by . We now get slightly more informal. The key point is that since none of the corners of the rectangle 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 and when it doesn’t should quickly convince you that the topology on will end up being the lower-limit topology.

Now, for the case with we have quite a different scenario. For example consider the anti-diagonal and consider for each fixed the basic open set clearly then from where it follows that 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 inherit the discrete topology as subspaces. For horizontal and vertical lines they still have the lower limit topology.

**9.**

**Problem:** Show that the dictionary order topology on the set is the same as the product topology on where denotes with the discrete topology. Compare this topology to the usual topology on

**Proof:** Let and let be a basic open neighborhood of it. Then, consider the set clearly this is open in with the lexicographic ordering and if we must have that which implies that and thus . Thus, noting finishes that portion of the argument. Conversely, let and be some basic open neighborhood containing it. If then and we’re done. So, assume that . If then consider . Clearly . Now, to see that we note that and thus and since we automatically have that .

Now, if then choosing and applying similar techniques works.

Lastly, if then choosing automatically works since and if we have that and . Regardless, we’ve found a basic open set in which contains and which is contained in .

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

The order topology on is finer than that of the usual topology on . Just note that if and is an open ball centered at it that is a basic open set in containing and which is contained inside the ball. The converse is not true. For example consider the origin and the basic open neighborhood . Any open ball centered at zero will contain points not on that line from where the conclusion follows.

**10.**

**Problem: **Let . Compare the product topology on , the dictionary order topology on , and the topology inherits as a subspace of in the dictionary order topology.

**Proof: **Let denote the topologies in the order they were mentioned. We first claim that . To see this let be arbitrary and note that generates . Then, if we have that . To see why the inclusion’s strict we note as in the last example that for example and any open square must intersect points of not in that interval.

Clearly then as was shown in the book.

No comments yet.

## Leave a Reply