I am confused in definition of disconnected set in topological space $X$.

Which of the following definition is valid/correct?

1) a subset $A$ of $X$ is disconnected if $A=V ∪W$ where $U$ and $V$ are non-empty disjoint open sets in $X$

2)a subset $A$ of $X$ is disconnected if $A⊆ V ∪W$ where $U$ and $V$ are non-empty disjoint open sets in $X$

Further I saw, an answer on MSE that, $\mathbb{Q}×[0,1]$ is not connected subspace in $\mathbb{R^2}$ with usual topology! the answer is like this!!

" As for any $r∈\mathbb{R}-\mathbb{Q}$ we can divide, given subspace into two nonempty disjoint open sets $(-∞,r)×[0,1]$ and $(r,∞)×[0,1]$ hence $\mathbb{Q}×[0,1]$ is not connected in $\mathbb{R^2}$ "

What they mean by "divide"? Please help me....

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    $\begingroup$ 2 is false. Just take the open unit interval as $A$ and also as $U$. Now you can take $V$ to be $(2,3)$. It becomes true if you require $A\cap U\neq\varnothing$ And $A\cap V\neq\varnothing$. $\endgroup$ – Clayton Mar 19 '18 at 4:23
  • $\begingroup$ @Clayton sir, then how $Q×[0,1]$ is disconnected and how $Q×[0,1]=(-∞,r)×[0,1]$ ∪$(r,∞)×[0,1]$ ? Where $r∈\mathbb{R}-\mathbb{Q}$ $\endgroup$ – Akash Patalwanshi Mar 19 '18 at 4:28

The first definition is correct. The second one cannot be correct. We see this by letting $A = V$. Then $A$ would be a connected component of a disconnected space, but $A$ itself is still connected. When considering connected spaces, we want to capture the notion of some "divide" or "void" between connected components. In the language of open sets, we phrase this as being able to "divide" the space into a union of disjoint open sets. In other words, "divide" just means that you are able to find such a union between disjoint open sets.

  • $\begingroup$ Sir, then how $Q×[0,1]=(-∞,r)×[0,1]$ ∪$(r,∞)×[0,1]$ ? Where $r∈\mathbb{R}-\mathbb{Q}$ $\endgroup$ – Akash Patalwanshi Mar 19 '18 at 4:26
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    $\begingroup$ You are right. It should be $\Bbb{Q} \cap (-\infty, r) \times [0,1] \cup \Bbb{Q} \cap (r, \infty) \times [0,1]$. $\endgroup$ – gian Mar 19 '18 at 4:31
  • $\begingroup$ Thank you so much sir :-) $\endgroup$ – Akash Patalwanshi Mar 19 '18 at 4:32
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    $\begingroup$ Also, you could save yourself some time by noting that $\Bbb{Q}$ is totally disconnected. Thus, any product space involving $\Bbb{Q}$ will also be disconnected. $\endgroup$ – gian Mar 19 '18 at 4:33
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    $\begingroup$ @AkashPatalwanshi, let's forget about $[0,1]$ for a second. $\Bbb{Q}$ is already disconnected by the following: $\Bbb{Q} \cap (-\infty, r) \cup \Bbb{Q} \cap (r, \infty)$, where $r$ is irrational. $\Bbb{Q}$ intersected with any open interval from $\Bbb{R}$ must also be open, assuming that $\Bbb{Q}$ inherits the subspace topology from $\Bbb{R}$. $\endgroup$ – gian Mar 19 '18 at 13:28

Let $T_X$ be the topology on $X.$ The subspace topology $T_A$ on $A\subset X$ is $T_A=\{t\cap A:t\in T\}.$

Any space is disconnected iff it is the union of two disjoint non-empty open sets.

So the sub$space$ $A$ is disconnected iff $A=B\cup C$ where $B,C$ are disjoint non-empty members of $T_A.$

Equivalently, the subspace $A$ is disconnected iff there exist $U,V\in T_X$ such that $U\cap V\supset X$ and such that $U\cap A, V\cap A$ are non-empty and disjoint.

Neither def'n 1) nor def'n 2) is correct.

  • $\begingroup$ Sir, That is we have check disconnectedness with respect to subspace topology? $\endgroup$ – Akash Patalwanshi Mar 19 '18 at 11:17
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    $\begingroup$ Yes. Connectedness or lack of it is a topological property. Without a specified topology it is not definable. $\endgroup$ – DanielWainfleet Mar 19 '18 at 11:20
  • $\begingroup$ Sir, then please then can you tell me how to show $\mathbb{Q}×[0,1]$ is disconnected as subspace of $\mathbb{R^2}$ $\endgroup$ – Akash Patalwanshi Mar 19 '18 at 11:21
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    $\begingroup$ Let $A=\Bbb Q\times [0,1].$ Let $U= (-\infty, \sqrt 2\;)\times \Bbb R$ and $V=(\sqrt 2\;,\infty)\times \Bbb R.$ Then $U,V$ are open in $\Bbb R^2.$ So $A\cap U, A\cap V$ are disjoint non-empty open subsets of the space $ A,$ and their union is $A.$ $\endgroup$ – DanielWainfleet Mar 19 '18 at 16:17

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