## 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.

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