## Munkres Chapter 2 Section 1

This begins a substantial effort to complete all of (except the first chapter) the problems in James Munkre’s Topology I have chosen to start at chapter 2 considering the first chapter is nothing but prerequisites. Thus, without a minute to spare let us being.

**1.**

**Problem: **Let be a topological space, let be a subset of . Suppose that for each there is an open set containing such that . Prove that is open.

**Proof:** We claim that but this is obvious since for each we have that . Conversely, since each we have that the union of all of them is contained in , namely . Thus, is the union of open sets and thus open.

**2. **

**Problem:** Compare the nine topologies on given in example 1.

**Solution: **This is simple.

**3.**

**Problem: **Show that given a set if we denote to be cocountable topology (a set is open if it’s complement is countable or the full space) that is a topological space. Is it still a topological space if we let ?

**Proof:** Clearly for the first part . Now, if is a collection of open sets then we note that is finite and thus for any . Thus, is finite and thus in . Now, if we have that and thus is the finite union of finite sets and thus finite, so .

If we redefine the topology as described it is not necessarily a topology. For example, give that topology and note that is a collection of elements of but whose complement is neither infinite, empty, or the full space. Thus, this isn’t a topology.

**4.**

**Problem:**

**a) **If is a family of topologies on , show that is a topology on . Is ?

**b) **Let be a family of topologies on . Show that there is unique topology on containing all the collections , and a unique largest topology contained in all of the .

**c) **If , let and . Find the smallest topology containing and the larges topology contained in .

**Proof:**

**a) ** Let and let be arbitrary. Then, by assumption we have that for every but since this was assumed to be a topology we have that for every that and thus . Now, if we must have that for every and thus for every and thus . Thus, noting that for every we must have that the conclusion follows.

No, the union of two topologies needn’t be a topology. Let be defined as in part c and note that but

**b) **This follows immediately from part a. For the first part let (where is the set of all topologies) and let , this clearly satisfies the conditions. For the second one merely take

**c) **We (as pointed out very poignantly in the previous problem) just take the union of the two topologies. So, we claim that the desired smallest topology is in fact . But this is just computation and we leave it to the reader. For the second part just intersect the two topologies.

**5.**

**Problem: **Show that if is a base for a topology on , then the topology generated by equals the intersection of all topologies on which contain . Prove the same if is a subbase

**Proof:** Let be the intersection of all topologies on which contain and the topology generated by . Clearly since is itself a topology on containing . Conversely, let we show that where is any topology on containing . But, this is obvious since for some and thus is the union of open sets in and thus in . The conclusion follows.

Next, let be above except now is a subbase. For the same reasons as above we have that . Conversely, for any topology containing we have that if then where each is the finite union of elements of . But, by construction it follows that each is open (it is the finite intersection of open sets in ) and thus is the union of open sets in and thus .

**6. **

**Problem:** Show that the topologies on and the Sorgenfrey line aren’t comparable

**Proof:** See the last part of the next problem

**7.**

**Problem: **Consider the following topologies on:

For each determine which of the others contain it.

**Solution:**

:Clearly we have that since the defining open base for is contained entirely in .

: But, since but and thus not in . Now, to prove the inclusion indicated we know that for each open set in the cofinite topology we have that and so if we assume WLOG that then

And thus is open in the usual topology.

: But, . To see this it suffices to show that is open in since this is a base for . But, to see this we must merely note that .

: Lastly, . To see this we must merely note that .

For the result is obvious except possibly how it relates to . But, in fact they aren’t comparable. To see this we first show that is not open in . To see this we show it can’t be written as the union of sets of the form and but it clearly suffices to do this for the latter sets. Now, to see that can’t be written as the union of sets of the form we recall from basic real number topology that is not open ( is not an interior point) and thus it can’t be written as the union of intervals or it would be open in the usual topology on . Also, consider .

**8.**

**Problem:** Show that the countable collection is a base for the usual topology on

**Proof:** This follows from the density of . It suffices to show that given any and any that there is some element such that . But, from basic analysis we know there is some rational number such that and similarly there is some such that . Thus, . The conclusion follows. .

**9.**

**Problem: **Show that the collection generates a different topology from the one on (the lower limit topology)$.

**Proof:** Clearly is open in but we show that it can’t be written as the union of elements of . So, suppose that where . Then, there exists some such that . Now, since we must have that but and thus and thus . Contradiction.

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