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Suppose $S$ is the union of two disjoint circles; consider each circle as one subset, there is no path from one subset to another one. Is there any special terminology for that?

Andrés E. Caicedo
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Amin
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    Partition maybe? – Bram28 Jun 25 '17 at 05:05
  • I think you are referring to set $S$ being partitioned into its respective equivalence classes. – Soby Jun 25 '17 at 05:05
  • I mean the "name" of such kind of sets not description; we say $S$ is .... . – Amin Jun 25 '17 at 05:14
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    The new edit is a totally different question to the original. You should revert this question and ask another. However, the answer is that the space is not path connected. See https://en.wikipedia.org/wiki/Connected_space#Path_connectedness – Evan Rosica Jun 25 '17 at 06:23
  • @EvanRosica Path-connected; this is the case. Thanks. – Amin Jun 25 '17 at 06:37
  • You're welcome. I've added this to my answer. If it was helpful feel free to accept it. – Evan Rosica Jun 25 '17 at 06:40

2 Answers2

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To answer the edited question:

We would say that the set is not path connected.

To answer the question as originally posed:

All sets can be written as a union of disjoint subsets. Say $S= \{s_1 ,s_2, \cdots , s_n \} $. Then $S$ is the union of the collection of all sets which are composed of exactly one element of $S$. More concisely, $S$ is the union of all its singleton subsets. Since any set has the property you described, the answer is simply a set.

For example, if $S= \{s_1 ,s_2, s_3 \}$, then $\cup \{ \{s_1 \} ,\{s_2\} ,\{s_3\}\}=\{s_1 ,s_2, s_3 \}=S$

Evan Rosica
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  • Consider a square. Is it possible to write it as the union of its closed subsets? I edited the question. – Amin Jun 25 '17 at 06:16
  • @Amin this is a different question. Adding that the subsets must be closed means that the answer depends on the topology since closure is a topological property. If the topology is such that singleton sets are closed, then my answer stands. – Evan Rosica Jun 25 '17 at 06:22
  • This was in my mind; I knew your answer and that is the trivial case. I talk not about discrete case. Not necessarily closed. – Amin Jun 25 '17 at 06:24
  • My answer holds for any $T_1$ space, not just the discrete case. See here https://math.stackexchange.com/questions/17649/are-singleton-sets-in-mathbbr-both-closed-and-open – Evan Rosica Jun 25 '17 at 06:25
  • What do you mean by $T_1$? – Amin Jun 25 '17 at 06:28
  • @Amin Its a class of topological spaces in which points (singleton sets) are closed. https://en.wikipedia.org/wiki/T1_space. See Jonas Meyer's answer in https://math.stackexchange.com/questions/17649/are-singleton-sets-in-mathbbr-both-closed-and-open. – Evan Rosica Jun 25 '17 at 06:29
  • you may remove the first part of the answer. – Amin Jun 25 '17 at 06:43
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Yes, it is called a union of pairwise disjoint sets. Introduce
your own term but be sure to include a definition when you use it.

William Elliot
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