## Homtopy and the Homotopy Category (Pt. I)

**Point of Post: **In this post we motivate, define, and discuss the notion of homotopy. We then introduce the homotopy category as a quotient category of .

**Motivation**

In topology we care about the geometry of a space–how we can describe all of the geometric properties that a space has. For example, we care if the space is compact, we care if it locally connected, we care if it . All of the properties factor into whether or not two given topological spaces are to be considered “the same”. That said, for large portions of mathematics some of the topological invariants of a space are less important than others. This can occur either because they literally matter less to us (for example, compactness is something that, while nice, isn’t an absolute necessity for a topological space to be nice). This can also be true because the subject area we are are working in contains spaces which necessarily already satisfy some of the properties (for example, metric spaces already satisfy all of the nice separation axioms).

## Path Connectedness

During our discussion of connectedness of topological groups I alluded to the fact that I would not be talking about path connectedness, but why? It seems natural that the two concepts, ostensibly so connected (), should be discussed at the same time. While I cannot refute that logic, my reason for separating (I’m on fire) them was purely a personal matter of connotation. To me connectedness while very geometric is much more aligned in the depths of point-set topology. It is a fundamental property of spaces. Path connectedness (while admittedly possessing all of the aforementioned properties) on the other hand has a *very* geometric flavor to it. Also, path connectedness will be an integral part of our discussion of manifolds. Thus, without further justification to a non-existent audience let us begin:

**Path:** Let be a topological space, then we call a continuous map such that and a *path* from to .

**Path Connected:** We call a topological space *path connected* if for any there exists a path between them.

Intuitively we can think about a path between two points as just connecting them with a “squiggly line”. So, it is natural to assume that every path connected space is connected but as some wise old guy once said “What is asserted without proof may be dismissed without proof”:

**Theorem:** Let be path connected, then is path connected.

**Proof:** Suppose that is a continuous surjection (where as always is the two-point discrete space). Then, choose . By assumption there exists a path between them. Clearly then, is a continuous surjection and so it follows that is also a continuous surjection, but this clearly contradicts the connectedness of .

So, what are some examples of path connected sets you may ask? We can find a plethora, but first we need a little itsy bitsy lemma (barely worth saying).

**Theorem:** Every convex subset of (a normed vector space over ) is path connected.

**Proof:** Clearly given any where is the convex set the map is continuous obviously and by assumption , from where the conclusion follows.

So, from this we can derive zillions and zillions (technical term of course) of path connected sets, for if you’ll remember we proved earlier that every open ball in a normed vector space is convex. So, in particular every open ball in is path connected. Also, realizing that themselves are convex (obviously) and in fact so is the punctured plane (just take a polygonal line if needed). Slightly less obvious is that for some arbitrary (once again take a polygonal line)

So, we begin to prove some pretty obvious theorems.

**Theorem:** Let be path connected and a continuous map. Then, is a path connected subspace.

**Proof:** Given it is evident that where is the path from to in is the required path.

**Theorem:** Let be a collection of path connected topological spaces, then with the product topology is path connected. The converse is also true.

**Proof:** Recall that if is a collection of continuous maps then the map

is connected (since each of the projections is connected). So, for any we merely take the guaranteed guaranteed path from to and note that

is the desired path.

Conversely, if is path connected then so are each of it’s coordinate spaces since the projection onto them is continuous.

**Theorem:** Let be path connected and a quotient space of , then is path connected.

**Proof:** This is blindingly obvious since the quotient map which defines the topology on is automatically a continuous surjection.

**Theorem:** If are two non-empty topological spaces, then is not path connected.

**Proof:** This obviously follows since we proved earlier that they aren’t even *connected*.

That said, there are some fairly neat theorems which are rooted very closely in geometric intuition. If for some fixed point we can connect it to any other point by a path we should be able to connect any two points by connecting the first point to , the second point to , and then “connecting” the paths. We formalize in the following paragraphs, but first we need a very important lemma.

**Theorem (Gluing Lemma):** Let be a topological space and a closed cover of it. Furthermore, suppose that for each there exists a continuous map such that for each . Then, there exists a unique continuous map such that .

**Proof:** First note that if then

and so the map

is at least well-defined, and in fact this map clearly satisfies . To see that it’s continuous we let be closed. Clearly

but each is a closed subset of a closed subspace of and thus closed in . Thus, is the finite union of closed sets and thus closed. Lastly, to see that is unique we merely note that if where another such map then given we have that for some and so

from where the conclusion follows.

**Corollary:** By an infinitesimal modification the above proves the same if is replaced by an arbitrary open cover

**Theorem (*):** Let be a topological space and suppose that for some fixed there is a path from to any other point of , then is path connected.

**Proof: **Let be arbitrary. By assumption there exists a path from to . Define a “reverse path” from to as follows

This is readily verified to be path from to . So, let be the path from to . Clearly the maps and are continuous and since

It follows from the gluing lemma that is continuous and so noting that

and

and so the theorem follows.

With this we have an easy way to prove another nice theorem

**Theorem:** Let be a topological space and a collection of path connected subspaces such that , then is path connectected.

**Proof:** Let , then given any we have that for some and so by assumption there exists some path from to . Clearly, if we extend the codomain we see that is also a path from to in . Considering our previous theorem finishes the argument.

With this we can prove a satisfying theorem, which in a very real sense is just mimicry of a previous theorem.

**Theorem:** Let be any normed vector space over . Then, is path connected.

**Proof:** We merely need to note that the map is continuous. Thus, by previous theorem is path connected. But,

and thus noting that

and appealing to the previous theorem finishes the argument.

We finish our discussion with a very, very nice theorem which in essence will show that every manifold is path connected. But, first we need to define the local analogue of path connectedness.

**Locally Path Connected:** Let be a topological space, then is said to be *locally path connected* if for each there exists a path connected neighborhood of it.

The role of local path connectedness plays a much more important role that it’s plain old connectedness analogue. This can be seen from the following theorem.

**Theorem:** Let be a locally path connected connected space, then is path connected.

**Proof:** Let and let be the set of all points in for which there is a path from to that point. Since () it suffices to prove that it is both open and closed (and the result will then follow from ‘s connectedness). So, let . By assumption there exists some neighborhood of it such that is path connected, we claim that . To see this, let . Using the exact same argument as in theorem (*) we can then construct a path from to and so . It follows that is open. Now, to show that is closed we show it is invariant under closure, and sine it suffices to show the reverse inclusion. So, let . Then, by ‘s local connectedness there exists a neighborhood of which is path connected. But, since there must be some point . So, using the exact same technique described in theorem (*) again we may connect the path from to and the path from to to get a path from to . It follows that .

Thus, is open and closed and since it’s non-empty it must be that . But, that means that every point of may be connected to by a path. Thus, by theorem (*) is path connected.

As stated this will prove that every manifold is path connected, but maybe a little closer to home is that this implies that *every* open connected set in is path connected, we state this in the following theorem.

**Theorem:** Let be a normed vector space and open and connected. Then, is path connected.

**Proof:** By assumption for each point there exists some open ball , but this open ball is convex and thus path connected. So, is locally path connected and connected thus by the previous theorem path connected. .

This confirms everyone’s intuition from back in complex analysis or calculus when it “seemed” that every connected “blob” in could have a “line drawn in it connecting two points”.

This ends our brief discussion of path connectedness.

## Thoughts about connectedness (Locally connected spaces)

Just as with compactness there is a certain local connectedness property that spaces may posses without being themselves connected.

**Locally Connected:** Let be a topological space, then we call *locally connected *if for every and every neighborhood of there exists some neighborhood connected of such that .

*Remark:* It is apparent that this is equivalent that has a connected open base.

We have already said that every normed vector space with the usual topology is locally connected (since every open ball is convex and thus connected). Also, it is clear that our example is locally connected but not connected and that the rationals are neither connected nor locally connected. In fact, you can have a locally connected space which is totally disconnected, take any discrete space.

But, unlike local compactness which is implied by compactness local connectedness is not implies by connectedness as the following example shows.

**Example (Topologist’s Sine Curve):** Define by . So, define . It is clear that is closed since it is equal to which is the graph of a continuous function on a connected space. But it is fairly easy to see that is not locally connected.

We now prove some facts about locally connected spaces.

**Theorem:** Let be a locally connected space. If is an open subspace of , then each component of is open in . In particular each component of is open.

**Proof:** Let be a component in and let . Since is a neighborhood of and is locally connected there exists some connected neighborhood of contained in . But, since is a component we must have that . The conclusion follows. .

**Theorem: **Let be a topological space where all the components of every open subspace of are open, then is locally connected.

**Proof:** Let and let be any neighborhood of it. Since is an open subspace of it’s components are also open in . In particular, the component which contains is open in . Thus, we have found a connected neighborhood of contained in . The question is whether or not it is connected with respect to . But, since the topology of as a subspace of is the same as that with respect to there is nothing to prove.

**Theorem:** Let be a compact locally connected space. Then, has finitely many components.

**Proof:** Suppose not. Then we have that the class of components of is infinite. Since the formation of components is a partition we see in particular that covers and it admits no finite subcovering. But, is in fact an open cover of by previous theorem. It follows that is an open cover of which admits no finite subcover, which of course contradicts its compactness.

**Theorem:** Let be locally connected and open, surjective, and continuous. Then, is locally connected.

**Proof:** Let be arbitrary and any neighborhood of . Since is continuous we have that is a neighborhood of . Now, by ‘s local connectedness there then exists some connected neighborhood of such that . It follows that is connected and open and . The conclusion follows. .

**Corollary:** Let be an arbitrary collection of topological spaces and locally connected. Then, is locally connected for every .

We know answer the question of whether or not local connectedness is invariant under products.

**Theorem:** Let be a finite collection of locally connected spaces. Then, is locally connected.

**Proof:** Let be arbitrary and any neighborhood of . Clearly then is a neighborhood of for each . Using ‘s local connectedness we are furnished with some open connected such that . Clearly then, is a connected open subspace of such that . The conclusion follows. .

The general case for the product of arbitrarily many locally connected spaces need not be locally connected. With a little bit of unjustified knowledge we can see that under the discrete topology is locally connected but where is the Cantor set. Clearly the Cantor set is not locally connected though.

If we strengthen our conditions though we’re right as rain.

**Theorem:** Let be an arbitrary collection of connected and locally connected spaces. Then, is locally connected under the product topology.

**Proof:** Let be arbitrary and any neighborhood of . We of course may assume WLOG that is basic open. So, let be the finitely many indices such that . For each we have that is a neighborhood of . And thus, by ‘s local connectedness there exists a connected neighborhood such that . Clearly then if

Then is open and connected and . The conclusion follows. .

*Remark:* The necessity for each to be connected was used when we were able to conclude that the product of the ‘s was connected.

## Thoughts about connectedness (Pt. II)

In the last post we gave quite a few equivalent definitions of a connected space, we now prove some basic theorems relating to connectedness.

Perhaps the easiest to relate to is the following.

**Theorem:** Let be a subspace of then it is connected if and only if it is an interval.

Suppose that is connected but it is not an interval. Then, there exists some such that for some we have that but . Clearly then is a disconnection of . This is of course a contradiction.

Conversely, suppose that is an interval and is a disconnection of . Since we may choose and and we may clearly assume that . Since is an interval and each point in is either in or we define . Evidently we have that and so (since it is an interval). Since is closed in it is clear from basic real line topology that . Thus, we conclude that . Again, by basic real line topology we know for every such that , and since is closed we must have that . This contradicts the assumption that .

**Corollary:** Every subset such that is dense is not connected.

*Remark:* Note that the only properties that were used above is the idea of what an interval is and the L.U.B. property of . It is not surprising then that this extends to the idea of a linear continuum (an order space with the L.U.B. property), but considering I haven’t defined and am not going to define this…there is no point in merely introducing this.

We next prove a mean feature of connectedness.

**Theorem:** Let be connected and be continuous, then is a connected subspace of .

**Proof:** Suppose not, and there exists some which are non-empty and open and such that and . Then we have that are non-empty disjoint open sets in such that . This obviously contradicts ‘s connectedness. The conclusion follows

This is the standard proof, but there is a slightly more eloquent one. (it is essentially the same, but worded differently).

**Proof:** For the sake of notational convenience let us assume that . Now, let us assume that is not connected. Then, there exists some which is continuous and surjective. Clearly then is continuous and surjective. This is a contradiction

**Corollary:** If then is connected if and only if is .

From this we can derive one of the most widely used theorems in calculus and analysis.

**Theorem (Generalized Intermediate Value Theorem-IVT): **Let be a connected space and be continuous. Then is an interval. In particular, given any such that there exists such that there exists some such that .

**Proof:** This follows from the fact that is a connected subspace of and thus an interval. .

**Corollary:** If is continuous then given any between and then there exists some such that .

The next is a neat other theorem consequent of the generalized mean value theorem.

**Theorem:** Let be connected and suppose there exists some non-constant continuous map then is uncountably infinite.

**Lemma: **Any interval of the form and are equipotent.

**Proof:** Define by then is easily verified to be a bijection. Now, consider by . This is equally easily verified to be a bijection. Thus, is a bijection. The conclusion follows

Now by assumption we know that is connected and continuous, and so by previous comment is an interval. But, by the assumption that is non-constant we know that it is an interval of either the form or . Since these two are equipotent we may assume WLOG that . Thus, by previous comment this implies that . It follows that there exists some which is a bijection. Now, if we assume that is countable then there exists some which is bijective. It follows that is a surjection, but this is clearly absurd. The conclusion follows

The next two theorems are just consequences of something I have forgot to say in general.

**Theorem:** Let be connected and a continuous map. Then, is connected.

**Proof:** As was proved earlier the map given by is continuous. Also, it is quite clear that and so is the continuous image of a connected set, and thus connected.

**Theorem:** Let be a connected space, then is connected.

**Proof: **The map given by is trivially continuous and it is clear that and so is the continuous image of a connected set, and thus connected. .

**Theorem:** Suppose that is an arbitrary collection of non-empty topological space. Then is connected implies that is connected for each .

**Proof:** Clearly we have that (the canonical projection) is continuous and surjective. Following the same logic as the previous problems we may conclude.

We now prove the converse of the above.

**Theorem:** Suppose that is an arbitrary collection of non-empty connected space, then is connected under the product topology.

**Proof: **Assume not, then there exists some (where is the two-point discrete space) which is continuous and non-constant. So, let be a fixed element of and consider some . Define by where

Clearly is continuous , and so is continuous. But, by the assumption that is continuous this means that is constant, let’s assume everything maps to . It follows that for every such that for every . Repeating this process with a different index we arrive at another set of such that they differ from in all but finitely many coordinates, which also map under to zero. Doing this for each we arrive at a dense subset of which all map to zero under . It follows that the zero function and are both continuous mappings into a Hausdorff space such that their agreement set is dense. It follows from a previous problem that . This contradicts the assumption that was constant. The conclusion follows

Before we make a corollary about this we prove something that everyone reading this (if there is anyone) knows but I haven’t actually taken the time to write down. Namely:

**Theorem:** under the usual metric topologies.

**Proof:** We use the canonical mapping given by . It is easily verified that is bijective. And in fact we see that and so it’s an isometry, thus clearly bicontinuous. The conclusion follows.

We are now prepared to state that

**Corollary:** and so are connected.

## Thoughts about connectedness

In this post we begin a series of discussions about one of the easiest to define concepts in topology. That said, it is one of the most important. We begin with some motivation.

If you were a calc II student and asked to visualize these two sets and and say what is the most obvious difference between them you most likely respond “The first one is in *one piece *and the second one isn’t”. This idea of something being in *one piece* is a fundamental concept. It intuitively fits with our idea of homeomorphisms that two spaces cannot be homeomorphic of one space is in two pieces and the other one.

There is also the vague notion of “reachability”. Imagine you took two people and placed one on a random spot in and the other in a random spot in . Then, the first person could walk to any other point of the set he’s on, while the other cannot. This idea of when two things can be traveled between without leaps or jumps is a huge concept in more advanced topology, that we will eventually hit on.

So we want to formalize this idea of being in *one piece*. Using our original example again we may be able to model this idea, which is commonly called *connectedness.*

**Connected: Let be a topological space. Then, is called connected if it cannot be expressed as where are open, non-empty and disjoint. **

We begin by proving a couple of equivalent definitions.

**Theorem:** Let be a topological space. Then, is connected if and only if the only sets which are simultaneously open and closed are and .

**Proof:**

Suppose that is connected, but there existed some which is both open and closed. Clearly then is both open and closed. Thus, are disjoint non-empty open sets such that . This is a contradiction.

Conversely, suppose that the only sets which are both open and closed are and . Next suppose that were disjoint non-empty open sets such that . Then since (this is easy to check) and is open it follows that is closed. But, since and it follows that and in particular . Thus, is a set that is both open and closed which is neither or . Contradiction.

**Theorem:** Let be a topological space. Then is both open and closed if and only if .

**Proof:**

Suppose that is both open and closed but . Then, let . We have two choices, if then every neighborhood of intersects and so but since it follows that , contradiction. If then every neighborhood of must contain a point of different from and so , but since is closed we know that and so . Contradiction. It follows that

Conversely, suppose that . Then, clearly and so from basic topology is closed. But, let then since there exists some neighborhood of such that does not intersect both . But since every neighborhood contains itself it follows that there exists some neighborhood of such that . It follows that and appealing once again to basic topology we see that is open. .

**Corollary:** A topological space is connected if and only if the only sets with empty interior are and .

Before we continue with need a definition.

**Separated sets:** If is a topological space and are such that we call *separated set.*

**Theorem:** A topological space is connected if and only if it cannot be written as the union of none-empty separated sets.

**Proof:**

Suppose that is connected but where are separated and non-empty. Clearly, and since it follows that . But, and so and so is open. But, it is also true that and that and so and so is open. Thus, , and are both non-empty and open. Lastly noting that we arrive at our contradiction.

Conversely, notice that if are open and disjoint then (this is easy to prove). Thus, if are disjoint non-empty open sets, then they are disjoint separated sets and thus . The conclusion follows.

This last one gives, in my opinion, the most elegant way of thinking about connectedness.

**Theorem:** If is a topological space, then it is connected if and only if the only continuous where with the discrete topology is a constant map.

**Proof:**

Suppose that is connected but is non-constant. Clearly then are open and non-empty. They are also clearly disjoint, for if then

Thus, are disjoint non-empty open sets and

This of course contradicts that is connected.

Conversely, suppose that the only which is continuous is a constant map, but are non-empty disjoint open sets such that . Define by

.

It is clear by the assumption that these two sets are non-empty, disjoint, and their union equals that this mapping is well define. Also,

.

Thus, is continuous. It follows that is a non-constant continuous map from to . This contradicts our assumption.

*Remark:* Clearly every two point discrete space is homeomorphic, so the usage of opposed to or is merely notational. Also, notice, by virtue that , that we could have rephrased the above as saying “A space is connected if and only if every continuous map from to is not surjective”

## Thoughts About Connectedness (Components)

It is not surprising that if one cannot speak of a connected space that they would like to at least speak of the “biggest” connected subspaces. To motivate our point let us consider our example we used in the first post. . It is intuitively clear once again what a component of this space would be, but defining it is a little trickier than may seem. We clearly don’t want to say a component is a connected subspace, for clearly is a connected subspace but not a “component”. It is clear once again that our minds are drawn to maximality. We want to describe our space by as few connected subspaces as possible. Defining this maximality is what requires finesse though. Usually in math when we think of biggest thing that contains something we usually think of the intersection of all things containing it. That is a problem here though, since the intersection of two connected sets need not be connected (think about the intersection of a line and a circle)….

With this in our noodles as motivation we begin the actual work.

**Component:** If is a topological space we define to be a *component* if it is connected and it is not contained in any other connected subspace.

Before we get into the meat of the argument we prove some facts regarding unions of connected sets.

**Theorem:** Let be a topological space and and arbitrary collection of non-empty connected subspaces. Then if is non-empty then is connected.

**Proof: **Suppose not, then there exists some which are open and non-empty and such that and and . Now, clearly no one particular may be in both otherwise so by assumption that we may conclude in particular that there exist some such that , and there is some . But, since we see that is a disconnection of and this is a contradiction.

We can actually prove some pretty cool theorems using this. For example:

**Theorem: **Let be a normed vector space over . Then is connected.

**Proof:** Fix and define by .

**Lemma:** is continuous.

**Proof:** Since these are both metric spaces we use the more metric space oriented definition of continuity.If this is trivial, so assume not. Let be given, then choosing such that we see that . The conclusion follows.

Now clearly is a connected subspace of since it is the continuous image of a connected space. But, seeing that and appealing to the previous theorems finishes the argument.

**Corollary: **Every Banach space is connected.

**Corollary:** and thus are connected.

**Corollary: ** with the norm is connected for any topological space .

We next wish to prove something that might have been intuitively obvious.

**Theorem:** Let be a topological space and be a connected subspace then given any we have that is connected, in particular is connected.

**Proof:** Suppose that is a disconnection of in (that is there restrictions to is a disconnection of in the normal sense). Since is connected we clearly must have that either or . Now, it is relatively easy to show that this implies that and thus which contradicts our choice of .

*Remark: *There is a particularly nice way of proving that is connected implies that is connected. To see this consider then from basic topology we have that and since must be either or which are both closed in it follows that . From where the conclusion is immediate.

**Corollary:** If is a topological space with a connected dense subset then is connected.

We can now prove the main properties of components which interest us.

**Theorem: **If is a topological space and then it is contained in exactly one component.

**Proof:** Define to be the set of all connected subsets of which contain . Clearly then and so by a previous theorem we see that is connected. This is clearly a component which contains Also, it is clearly the only one. For, suppose that is a component which contains then and so and by ‘s maximality it follows that . The conclusion follows. .

**Theorem:** Any connected subspace of is contained in exactly one component.

**Proof:** The fact that if they are contained in one it is unique is obvious. To see that it is contained in at least one merely note that it is contained in the component constructed in the above theorem. .

**Theorem: **The components of are closed.

**Proof:** Let be a component of , then by previous theorem we have that is a connected subspace of which contains , and by ‘s maximality it follows that . .

**Theorem:** A connected subspace of which is both open and closed is a component of .

**Proof:** Suppose that is such a subspace and where is another connected subspace of . Then is clearly a disconnection of . It follows that no connected subspace of may properly contain . The conclusion follows. .

One of the important facts about components is the following.

**Theorem: **Let be a topological space and the set of components of , then forms a partition of .

**Proof:** Since every point of belongs to at least one component it’s clear that . Also, suppose that were such that then by a previous theorem we have that is a connected subspace of which properly contains and , which of course contradicts their maximality. The conclusion follows.

**Corollary:** If is a component of then

From this we can prove the following theorem.

**Theorem:** Let have finitely many components, then they are open subsets of .

**Proof:** Let be as above and let then which is the finite union of closed subspaces of and thus a closed subspace. The conclusion follows.

This finished up what I wanted to say about components. So, I am just going to prove some related theorems. Most of them relate to union of connected subspaces.

**Theorem:** Let be a topological space and a sequence of closed subspaces of such that . Then is a connected subspace of .

**Proof:** Suppose there existed non-empty open subsets of such that their restrictions to form a disconnection. Let . Clearly we must have that either or (it obviously can’t have some in one and some in the other or the restrictions of to would be a disconnection). WLOG assume that and let . If we are done, so assume not. Then by the well-ordering principle exists. Clearly we must have that .

Now, assume that then since we have that are both non-empty open subspaces of and their union is clearly . This of course contradicts ‘s connectedness. It follows that which contradicts the minimality of . It follows that . The conclusion follows.

**Theorem:** Let be a topological space and a collection of connected subspaces of such that . Then, is a connected subspace of .

**Proof:** Suppose not, and there existed some were are non-empty and open in and is a disconnection of . Then, since are non-empty and each connected we must have that and for some but since it is pretty plain that is a disconnection of . The fact that is disconnected is therefore untenable. .

We now give an alternate theorem that the product of finitely many connected spaces is connected.

**Theorem:** Let and be connected topological spaces, then is connected under the product topology.

**Proof: **Fix and . Clearly for any and . Thus, given any we have that is the union of two interesting connected sets, thus connected. Also, we have that for any . It follows that is connected. .

**Corollary:** It follows from induction that the product of finitely many connected spaces is connected.

We now need a definition. Note though that the following may be extended considerably.

**Convex set:** Let be a normed vector space over . We then call *convex *if for every .

**Theorem:** Let be a normed vector space and a convex subset, then is connected.

**Proof:**

**Lemma: **Let and define . Then if the mapping given by is continuous. (it is assumed that is a subspace of under the usual topology)

**Proof:** Notice that if , that

Now it is assume that we see that . So, letting we see that

The conclusion follows.

Now that would help us if we knew that is connected, so we must prove it. Don’t fret though, we must merely note that if is given by then it is evident that is continuous and . It follows from an earlier question that and thus is connected.

So, assume that is continuous and surjective. Let then clearly the mapping is also continuous and surjective. This contradicts that is connected. The conclusion follows.

From this we can derive a theorem which will become mainly interesting to us in the next post.

**Theorem:** Let be a normed vector space. Then the open (or closed) ball is convex for every and for every .

**Proof:** Let then

it follows that . The conclusion follows. .

**Corollary:** Every open (or closed) ball in a normed vector space is connected.