Abstract Nonsense

Crushing one theorem at a time

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 X be a subspace of \mathbb{R} then it is connected if and only if it is an interval.

Suppose that X\subseteq \mathbb{R} is connected but it is not an interval. Then, there exists some y\in \mathbb{R} such that for some x,z\in X we have that x<y<z but y\notin X. Clearly then \left(X\cap\left(-\infty,y\right)\right)\cup\left(X\cap\left(y,\infty\right)\right) is a disconnection of X. This is of course a contradiction.

Conversely, suppose that I\subseteq\mathbb{R} is an interval and A\cup B is a disconnection of I. Since A,B\ne\varnothing we may choose a\in A and b\in B and we may clearly assume that a<b. Since [a,b] is an interval and each point in [a,b] is either in A or B we define \gamma=\sup\left\{x:x\in A\cap [a,b]\right\}. Evidently we have that a\leqslant \gamma\leqslant b and so \gamma\in X (since it is an interval). Since A is closed in X it is clear from basic real line topology that \gamma\in A. Thus, we conclude that \gamma<b. Again, by basic real line topology we know y+\varepsilon\in B for every \varepsilon>0 such that y+\varepsilon\leqslant z, and since B is closed we must have that \gamma\in B. This contradicts the assumption that A\cap B=\varnothing. \blacksquare

Corollary: Every subset E\subseteq\mathbb{R} such that E' 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 \mathbb{R}. 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 X be connected and \varphi:X\to Y be continuous, then \varphi(X) is a connected subspace of Y.

Proof: Suppose not, and there exists some G,H\subseteq Y which are non-empty and open and such that \left(G\cap\varphi(X)\right)\cap\left(H\cap\varphi(X)\right)=\varnothing and \left(G\cap\varphi(X)\right)\cup\left(H\cap\varphi(X)\right)=\varphi(X). Then we have that \varphi^{-1}\left(G\cap\varphi(X)\right)=\varphi^{-1}(G),\varphi^{-1}(G\cap\varphi(X))=\varphi^{-1}(H) are non-empty disjoint open sets in X such that \varphi^{-1}(G)\cup\varphi^{-1}(H)=\varphi^{-1}(G\cup H)=X. This obviously contradicts X‘s connectedness. The conclusion follows \blacksquare

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 Y=\varphi(X). Now, let us assume that \varphi(X) is not connected. Then, there exists some \psi:\varphi(X)\to D which is continuous and surjective. Clearly then \psi\circ\varphi:X\to D is continuous and surjective. This is a contradiction \blacksquare

Corollary: If X\approx Y then X is connected if and only if Y is .

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

Theorem (Generalized Intermediate Value Theorem-IVT): Let X be a connected space and \varphi:X\to \mathbb{R} be continuous. Then \varphi(X) is an interval. In particular, given any c\in\mathbb{R} such that there exists x,y\in X such that \varphi(x)<c<\varphi(y) there exists some c_0\in X such that \varphi(c_0)=\varphi(c).

Proof: This follows from the fact that \varphi(X) is a connected subspace of \mathbb{R} and thus an interval. \blacksquare.

Corollary: If \varphi:[a,b]\to\mathbb{R} is continuous then given any c between \varphi(a) and \varphi(b) then there exists some c_0\in[a,b] such that \varphi(c_0)=c.

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

Theorem: Let X be connected and suppose there exists some non-constant continuous map \varphi:X\to\mathbb{R} then X is uncountably infinite.

Lemma: Any interval of the form (a,b),\text{ }a<b and \mathbb{R} are equipotent.

Proof: Define f:(a,b)\to (-1,1) by x\mapsto \displaystyle \frac{2}{b-a}(x-a)-1 then f is easily verified to be a bijection. Now, consider w:(-1,1)\to\mathbb{R} by \displaystyle x\mapsto \frac{x}{1+|x|}. This is equally easily verified to be a bijection. Thus, w\circ f:(a,b)\to\mathbb{R} is a bijection. The conclusion follows \blacksquare

Now by assumption we know that X is connected and \varphi continuous, and so by previous comment \varphi(X) is an interval. But, by the assumption that \varphi is non-constant we know that it is an interval of either the form [a,b] or (a,b). Since these two are equipotent we may assume WLOG that \varphi(X)=(a,b),\text{ }a<b. Thus, by previous comment this implies that \varphi(X)\simeq\mathbb{R}. It follows that there exists some \eta:\varphi(X)\to\mathbb{R} which is a bijection. Now, if we assume that X is countable then there exists some \psi:\mathbb{N}\mapsto X which is bijective. It follows that \eta\circ\varphi\circ\psi:\mathbb{N}\to\mathbb{R} is a surjection, but this is clearly absurd. The conclusion follows \blacksquare

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

Theorem: Let X be connected and \varphi:X\to Y a continuous map. Then, \Gamma_\varphi\subseteq X\times Y is connected.

Proof: As was proved earlier the map \iota_X\oplus\varphi:X\to X\times Y given by x\mapsto(x,\varphi(x)) is continuous. Also, it is quite clear that \Gamma_\varphi=\left(\iota_X\oplus\varphi\right) and so \Gamma_\varphi is the continuous image of a connected set, and thus connected.

Theorem: Let X be a connected space, then \Delta_X\subseteq X\times X is connected.

Proof: The map  \iota_X\oplus \iota_X:X\to X\times X given by x\mapsto (x,x) is trivially continuous and it is clear that \Delta_X=\left(\iota_X\oplus\iota_X\right)\left(X\right) and so \Delta_X is the continuous image of a connected set, and thus connected. \blacksquare.

Theorem: Suppose that \left\{X_j\right\}_{j\in\mathcal{J}} is an arbitrary collection of non-empty topological space. Then \displaystyle \prod_{j\in\mathcal{J}}X_j is connected implies that X_\ell is connected for each \ell\in \mathcal{J}.

Proof: Clearly we have that \displaystyle \pi_\ell:\prod_{j\in\mathcal{J}}X_j\mapsto X_\ell (the canonical projection) is continuous and surjective. Following the same logic as the previous problems we may conclude. \blacksquare

We now prove the converse of the above.

Theorem: Suppose that \left\{X_j\right\}_{j\in\mathcal{J}} is an arbitrary collection of non-empty connected space, then \displaystyle X=\prod_{j\in\mathcal{J}}X_j is connected under the product topology.

Proof: Assume not, then there exists some f:X\to D (where D is the two-point discrete space) which is continuous and non-constant. So, let \displaystyle \bold{x}=\prod_{j\in\mathcal{J}}\{x_j\} be a fixed element of X and consider some \ell\in\mathcal{J}. Define w:X_\ell\to X by \displaystyle x\mapsto\prod_{j\in\mathcal{J}}\{y_j\} where

\displaystyle y_j=\begin{cases} x_j & \mbox{if} \quad j\ne \ell \\ x & \mbox{if} j=\ell \end{cases}

Clearly w is continuous , and so f\circ w:X_\ell\to D is continuous.  But, by the assumption that X_\ell is continuous this means that f\circ w is constant, let’s assume everything maps to 0. It follows that f(\bold{y})=0 for every \bold{y}\in X such that \pi_j(\bold{x})=\pi_j(\bold{x}) for every j\in\mathcal{J}-\{\ell\}. Repeating this process with a different index k\in\mathcal{J}-\{\ell\} we arrive at another set of \bold{y} such that they differ from \bold{x} in all but finitely many coordinates,  which also map under f to zero. Doing this for each j\in\mathcal{J} we arrive at a dense subset of X which all map to zero under f. It follows that the zero function 0:X\mapsto D and f are both continuous mappings into a Hausdorff space such that their agreement set A(0,f) is dense. It follows from a previous problem that f=0. This contradicts the assumption that f was constant. The conclusion follows \blacksquare

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: \mathbb{C}^n\approx\mathbb{R}^{2n} under the usual metric topologies.

Proof: We use the canonical mapping \alpha:\mathbb{C}^n\to\mathbb{R}^{2n} given by (a_1+ib_1,\cdots,a_{2n-1}+ib_{2n})\mapsto (a_1,b_1,\cdots,a_{2n-1},b_{2n-1}). It is easily verified that \alpha is bijective. And in fact we see that \|\alpha(z)\|=\|z\| and so it’s an isometry, thus clearly bicontinuous. The conclusion follows.

We are now prepared to state that

Corollary: \mathbb{R}^n and so \mathbb{C}^n are connected.

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March 28, 2010 - Posted by | General Topology, Topology, Uncategorized | , , ,

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