connected sets

Section 5.1 connected sets

The first new topological notion is that of a connected set. Intuitively, a connected set comes "all in one piece". But we want to convert the idea of being "in one piece" into terms that make sequential, metric, and topological sense.

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Definition 5.1.1.

A disconnection of a set \(A\) is a pair of open sets \(U_1\text{,}\) \(U_2\) so that:

\(A\cap U_1\) and \(A\cap U_2\) are nonempty, and

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\(A\subseteq U_1\cup U_2\text{,}\) and

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\(\displaystyle U_1\cap U_2=\varnothing\)

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If a disconnection of \(A\) exists, we say that \(A\) is disconnected; if no disconnection exists, we say that \(A\) is connected.

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Example 5.1.2.

The set \(\mathbb{Q}\subseteq\mathbb{R}\) admits a disconnection: \(U_1=(-\infty,\sqrt{2})\) and \(U_2=(\sqrt{2},\infty)\text{.}\)

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In fact, it’s clear that any irrational number \(r\) provides a disconnection of \(\mathbb{Q}\text{.}\)

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Establishing that a set is disconnected is straightforward: simply describe the disconnection. But to show that a set is connected is somewhat less obvious.

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Example 5.1.3.

\(\mathbb{R}\text{,}\) considered as a subset of itself, is connecteed. To see this, observe that a disconnection \(U_1,U_2\) would have \(U_1=U_2^c\) and \(U_2=U_1^c\text{.}\) Thus \(U_1\) and \(U_2\) would both be clopen sets. But we have seen that the only clopen subsets of \(\mathbb{R}\) are \(\mathbb{R}\) and \(\varnothing\text{,}\) so at least one of \(U_1,U_2\) must have been empty (a contradiction to Definition 5.1.1.

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Actually, pretty much any argument for a set’s connectedness will have to rely on a proof by contradiction -- because being connected is a negative idea: to be connected means to not admit any disconnections.

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Checkpoint 5.1.4.

The connected subsets of \(\mathbb{R}\) are precisely the intervals.

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Proposition 5.1.5.

A set \(A\) is connected if any only if, given any open set \(U\) which intersects \(A\text{,}\) either \(U\subseteq A\text{,}\) or \(A\setminus U\) is not open.

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Proof.

This is just rewriting Definition 5.1.1.

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Now we’re ready for our first theorem of calculus.

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Theorem 5.1.6. Bolzano’s Intermediate Value Theorem.

If \(f:[a,b]\to \mathbb{R}\) is a continuous function, and \(y\in\mathbb{R}\) lies between \(f(a)\) and \(f(b)\text{,}\) then there is \(c\in[a,b]\) so that \(f(c)=y\text{.}\)

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This will follow from Checkpoint 5.1.4 and this important topological fact:

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Proposition 5.1.7. Continuous functions preserve connectedness.

Let \(f:A\to W\) be a continuous function from a subset of one normed space to another normed space. If \(A\) is connected, then so is

\begin{equation*} f_*(A)=\left\{f(a)\middle\vert a\in A\right\}\ \ . \end{equation*}

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Proof.

Our job is to prove that \(f_*(A)\) is connected, so let’s formulate the contrapositive: if \(f_*(A)\) is disconnected, then so is \(A\text{.}\)

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Assuming \(f_*(A)\) is disconnected means there’s a disconnection \(U_1,U_2\) of \(f_*(A)\text{.}\) I claim that \(f^*(U_1),f^*(U_2)\) disconnects \(A\text{.}\)

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First: because \(f\) is continuous, we know that \(f^*(U_1)\) and \(f^*(U_2)\) are open. Moreover, any element in \(U_1\cap f(A)\) is an element \(w\in W\) with \(f(a)=w\) for some \(a\in A\text{.}\) So \(a\in f^*(U_1)\cap A\text{;}\) and similarly \(f^*(U_2)\) is nonempty.

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Since \(f^*(A)\subseteq U_1\cup U_2\text{,}\)

\begin{equation*} A\subseteq f^*(f_*(A))\subseteq f^*(U_1\cup U_2)=f^*(U_1)\cup f^*(U_2)\ \ . \end{equation*}

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Finally, any \(z\in f^*(U_1)\cap f^*(U_2)\) would have \(f(z)\in U_1\cap U_2\text{,}\) so we see that \(f^*(U_1)\) and \(f^*(U_2)\) are disjoint.

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Thus we’ve shown the contrapositive of our claim.

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Checkpoint 5.1.8.

The proof of Proposition 5.1.7 requires a lot of manipulations of \(f^*\) and \(f_*\text{.}\) Justify each of these steps.

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Checkpoint 5.1.9.

Prove this sequential characterization of disconnection: \(A\) is disconnected by a pair of disjoint sets \(U_1,U_2\) if and only if for any sequence \(x_n\) of points of \(A\) which converges to a point of \(A\text{,}\) eventually \(x_n\) lies in \(U_1\) or in \(U_2\text{.}\)

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Remark 5.1.10.

Bolzano’s Intermediate Value Theorem is going to turn out to be pretty useful to us. But it’s also important to realize Proposition 5.1.7 has a much further reach. Consider the following example.

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Example 5.1.11.

We’ll model the current temperature on the surface of the earth by a function \(T:S^2\to \mathbb{R}\text{,}\) where \(S^2\) is the unit sphere. It’s pretty reasonable to expect \(T\) to be a continuous function.

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Furthermore, \(S^2\) is connected (that requires proof but we’ll assume it for now). So the set of current temperatures \(T_*(S^2)\subseteq \mathbb{R}\) is connected. That means we get an intermediate temperature theorem: for any measured temperatures \(T(x_1)\text{,}\) \(T(x_2)\text{,}\) and any temperature \(y\) between them, there is a point on earth \(x_y\) with \(T(x_y)=y\text{.}\)

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But we can push it further. Each point \(x\) on the globe has an antipode, which we’ll denote \(-x\text{.}\) The assignment \(x\mapsto -x\) is continuous, so the function

\begin{align*} \Delta:S^2&\to \mathbb{R}\\ x&\mapsto T(x)-T(-x) \end{align*}

(which measures the difference between the current temperature at \(x\) and the current temperature at is antipode) is continuous. Therefore the image \(\Delta_*(S^2)\subseteq\mathbb{R}\) is connected, i.e., an interval. Notice that \(\Delta(-x)=-\Delta(x)\text{,}\) so if \(\Delta_*(S^2)\) has any positive values, it also has negative values, and vice versa. In any case, \(0\in \Delta_*(S^2)\text{.}\) But that means there is a point \(x_0\in S^2\) with \(\Delta(x_0)=0\), i.e., \(T(x_0)=T(-x_0)\text{.}\) That is to say,

There are always antipodes with the same temperature.
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