# Connected set

1. Feb 22, 2013

### Bachelier

Is a singleton a connected set.

I am thinking it is because it is not in the intersection of two disjoint sets.

2. Feb 22, 2013

### WannabeNewton

Yes. An equivalent condition is that a set is disconnected if there exists a non - empty proper clopen subset contained in that set. The only clopen subsets of a singleton are the singleton and the empty set.

3. Feb 22, 2013

### micromass

Staff Emeritus
I'm not sure what this has to do with connected...

4. Feb 22, 2013

### WannabeNewton

I think we can give him the benefit of the doubt and assume he meant it isn't the union of two disjoint non - empty open subsets. To the OP, for a little bit more fun try proving the "Topologists' Sine Curve" is connected but neither locally connected nor path connected: http://en.wikipedia.org/wiki/Topologist's_sine_curve. It is rather instructive. Cheers!

5. Feb 22, 2013

### micromass

Staff Emeritus
This is truly one of the most elegant counterexamples in topology.
If you finished that one, maybe you can try to find some variations yourself such as connected and path connected, but not locally connected.

6. Feb 22, 2013

### WannabeNewton

Indeed it is. Now if only I could figure out the dang long line...;)

7. Feb 22, 2013

### Bachelier

Yeah that is what I meant. I was using Rudin's 2.45 definition: "A set E is said to be connected if E is NOT the union of two non∅ separated sets...You mentioned the word open in your statement...must they be open?

Thanks

8. Feb 22, 2013

### Bachelier

Reading the example right now. BTW this is the curve drawn on the front of Charles Pugh textbook.

9. Feb 22, 2013

### Bachelier

BTW what is the difference between a Topological Space and a Metric Space?

10. Feb 22, 2013

### WannabeNewton

$X$ is connected $\Leftrightarrow X$ cannot be written as the union of two nonempty separated sets $\Leftrightarrow$ The only subsets of $X$ which are clopen are $X$ and the empty set $\Leftrightarrow$ $X$ cannot be written as the union of two disjoint nonempty open subsets.

11. Feb 22, 2013

### WannabeNewton

A topological space is simply a set $X$ together with a subset of $\wp (X)$ (power set of $X$) that satisfies three specific conditions (the set itself and the empty set must be in the topology, arbitrary unions of open sets must be open, finite intersections of open sets must be open). It is more general than a metric space which is a set $M$ together with a metric $d:M\times M\rightarrow \mathbb{R}$ that of course satisfies the usual three properties you already know from Rudin. The metric induces a topology on $M$ which is the topology generated by the open balls of this metric (we say this collection of open balls is a basis for the topology). Many of the metric space notions such as continuity of maps and convergence of sequences can be naturally generalized to topological spaces e.g. let $X,Y$ be topological spaces and $f:X\rightarrow Y$ a map. We say $f$ is continuous if $\forall V\subseteq Y$ open in $Y$, $f^{-1}(V)$ is open in $X$. You can show easily that in the case of metric spaces, this definition is equivalent to the epsilon delta one.

12. Feb 22, 2013

### Bachelier

I am trying to picture this definition geometrically (for instance in $\mathbb{R} × \mathbb{R}$)

Please check my attachment. Notice that $E = E_1 \bigcup E_2$ but neither proper subset is clopen.

Further even if we consider $E_1$ by itself which is connected, it is not open.

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13. Feb 22, 2013

### WannabeNewton

In the diagram, you state that both $E_{1},E_{2}$ are closed. Do you know the definition of a closed subset? We define $A\subseteq X$ to be closed if $X\setminus A$ is open in $X$. Therefore if $E = E_{1}\cup E_{2}$, $E_{1},E_{2}$ are closed, and $E_{1}\cap E_{2} = \varnothing$ we can easily conclude that $E \setminus E_{1} = E_{2}$ is open and $E \setminus E_{2} = E_{1}$ is also open so they are both clopen. Keep in mind that your sets $E, E_{1}, E_{2}$ are proper subsets of $\mathbb{R}^{2}$ therefore when detecting whether $E$ is connected or not using open sets you must do so with respect to the subspace topology on $E$. This answers your last point as well: every set is open in itself by definition of a topology.

14. Feb 22, 2013

### micromass

Staff Emeritus
Too add to wbn, the crucial point here is that of a "subspace". Of course $E_1$ and $E_2$ are not open in entire $\mathbb{R}^2$. But in order to see that the space $X=E_1\cup E_2$ is connected, we don't work in $\mathbb{R}^2$, but in the subspace $X$.

Now, whether $E_1$ is closed in the subspace $X$ depends entirely how we define the topology on $X$. Since you're comfortable with metrics, we can just restrict the metric from $\mathbb{R}^2$ to $X$.

So for $x,y\in X$, we set $d_X(x,y)=d_{\mathbb{R}^2}(x,y)$. Now $X$ forms a metric space. Can you see that $E_1$ is open in $X$?? Try the sequential definition of open maybe: can you see that if a sequence converges to a point in $E_1$, then the sequence has to be in $E_1$ eventually? The trick is that the sequence is not a sequence in $\mathbb{R}^2$, but in $X$. So the elements of the sequence are either in $E_1$ or $E_2$.

15. Feb 22, 2013

### Bachelier

yup, you are right. Using topological definitions of openness, closedness and connectedness is more lucid than Rudin's neighborhoods and limit points definitions.

16. Feb 22, 2013

### WannabeNewton

Every metric space is first countable so the sequence lemma applies which basically allows us to use sequences to characterize many topological properties such as closure, interior, openness, and closedness. It is a very handy tool in proofs because even though the basic topological definitions may look elegant, they are not always very easy to work with in proofs. Cheers!

17. Feb 25, 2013

### mathwonk

oddly enough, the somewhat unintuitive definition which follows is easily shown equivalent to other definitions, but much easier to use in proofs.

A space X is connected if and only if every continuous map f:X-->{0,1} is constant. Hence X is disconnected iff there is a continuous surjection X-->{0,1}.

Equivalently (exercise), a set X is disconnected iff X is the union of two non empty disjoint open sets.

Certainly every map on a one point set is constant.

For the topologists sine curve it is easy to show a continuous map f is constant on the curvy part (if you know that (0,1) is connected), and then f also has the same value on the point (0,0), since that is in the closure of the curvy part.

I invite you to try this version on any connectedness proof you like, such as the continuous image of a connected set is connected, or the closure of a connected set is connected, or path connected implies connected....

Also a union of connected sets with non empty intersection is connected. Also a finite union of connected sets A1,...An such that each of A1,...,An-1 meets the next one is connected. These take a little effort by the open set definition (even Dieudonne' spends half a page on these last two in Foundations of Modern Analysis), but are all trivial by the constant function definition.

Last edited: Feb 25, 2013