
#1
Feb2213, 01:07 AM

P: 376

Is a singleton a connected set.
I am thinking it is because it is not in the intersection of two disjoint sets. 



#2
Feb2213, 01:14 AM

C. Spirit
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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.




#4
Feb2213, 12:36 PM

C. Spirit
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Connected set 



#5
Feb2213, 12:44 PM

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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
Feb2213, 01:00 PM

C. Spirit
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#7
Feb2213, 10:57 PM

P: 376

Thanks 



#8
Feb2213, 10:59 PM

P: 376





#9
Feb2213, 11:04 PM

P: 376

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




#10
Feb2213, 11:09 PM

C. Spirit
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#11
Feb2213, 11:18 PM

C. Spirit
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P: 4,915





#12
Feb2213, 11:27 PM

P: 376

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. 



#13
Feb2213, 11:33 PM

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




#14
Feb2213, 11:41 PM

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P: 16,537

Now, whether [itex]E_1[/itex] is closed in the subspace [itex]X[/itex] depends entirely how we define the topology on [itex]X[/itex]. Since you're comfortable with metrics, we can just restrict the metric from [itex]\mathbb{R}^2[/itex] to [itex]X[/itex]. So for [itex]x,y\in X[/itex], we set [itex]d_X(x,y)=d_{\mathbb{R}^2}(x,y)[/itex]. Now [itex]X[/itex] forms a metric space. Can you see that [itex]E_1[/itex] is open in [itex]X[/itex]?? Try the sequential definition of open maybe: can you see that if a sequence converges to a point in [itex]E_1[/itex], then the sequence has to be in [itex]E_1[/itex] eventually? The trick is that the sequence is not a sequence in [itex]\mathbb{R}^2[/itex], but in [itex]X[/itex]. So the elements of the sequence are either in [itex]E_1[/itex] or [itex]E_2[/itex]. 



#15
Feb2213, 11:44 PM

P: 376





#16
Feb2213, 11:51 PM

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#17
Feb2513, 12:36 PM

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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,...,An1 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. 


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