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Clearly if I have an equivalent definition of some property, I can get equivalent results, but sometimes it´s better one point of view than the other. In this context a clear equivalent definition of this property is that there does not exist a continuous function $f\colon X\to\{0,1\}$, and two disjoint sets $A,B$ ( such that the union of them is all $X$) and $f(A ) = 0$ and $f(B) = 1.$ I know that this question is very "open" but, could be useful this definition to prove something, in a better way than by a separation?

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I edited your question a bit. I hope you don't mind. – Grumpy Parsnip Aug 30 '11 at 21:40
It's, I think, easier to say that X is connected iff all continuous functions from X into the discrete space of 2 points are constant. This is used as a starting point for other generalizations in categorical topology. It also makes some proofs nicer. – Henno Brandsma Aug 30 '11 at 22:38
@Henno please can you link me some proof of that? – Daniel Aug 30 '11 at 22:51
Consider a non-constant f:X-->{0,1}. Then {1},{0} are both open in {0,1}, and $f^{-1}({1})$=U and $f^{-1}({0})$ are disjoint open sets whose union is X, and X is then disconnected. – gary Aug 30 '11 at 22:58
up vote 1 down vote accepted

As you point out, usually a property can be defined in various equivalent ways. Sometimes the equivalence of different formulations is not obvious at all. It might even be a deep theorem. In such a case it may be crucial to choose the relevant formulation when proving a theorem.

However, sometimes the equivalence is very easy to see or almost trivial as in your case of defining disconnectedness in two ways: if you have the separation $X=A\cup B$, you can define a continuous function $f:X\to \{0,1\}$ by $f[A]=\{0\}$ and $f[B]=\{1\}$; conversely if you have the continuous function $f$, then the sets $f^{-1}\{0\}$ and $f^{-1}\{1\}$ give a separation. This means that it does not matter very much which formulation you use when proving theorems about connected/disconnected spaces, since you can do the above easy deductions inside the proof if needed.

Trying to think of a case when the functional definition of disconnectedness is (a little bit) more useful, I could not come up with anything but the following (perhaps artificial) example.

Assume you have a family $\{X_i : i\in I\}$ of disconnected spaces. Then by the functional definition of disconnectedness you get, for each $i\in I$, nonempty disjoint sets $A_i,B_i$ with $A_i\cup B_i=X_i$ and a continuous surjective function $f_i:X_i \to \{0,1\}$ with $f[A_i]=\{0\}$ and $f[B_i]=\{1\}$. Now consider the function $f:\prod_{i\in I}X_i\to \{0,1\}^I$ defined by $f((x_i)_{i\in I})=(f_i(x_i))_{i\in I}$. This function $f$ is also continuous and surjective since $f_i,i\in I$ are. Hence the space $\{0,1\}^I$ is a continuous image of $\prod_{i\in I}X_i$. In particular, the Cantor set $\{0,1\}^\mathbb{N}$ (and thus any compact metrisable space) is a continuous image of any countably infinite (or larger) product of disconnected spaces.

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Let $X$ be a topological space. The following conditions are equivalent:

  1. If $U,V \subset X$ are disjoint open sets such that $U \cup V = X$ and then $U$ or $V$ is empty.
  2. The only clopen subsets of $X$ are $X$ and the empty set.
  3. Every continuous function $f: X \rightarrow \{0,1\}$ is constant where $\{0,1\}$ is endowed with the discrete topology.

I think that 1 and 2 are equivalent is pretty straightforward. For the equivalence of 1 and 3 proving that 1 implies 3 you might as well prove that connected sets are preserved by continuous functions and the connected components of $\{0,1\}$ are $\{0\}$ and $\{1\}$. Proving that 3 implies 1 is also pretty simple as per Brian's suggestion if you have two disjoint open subsets of $X$ such that $U\cup V=X$ then if we define $F: X \rightarrow \{0,1\}$ as $f(U)=0$ and $f(V)=1$ this must be constant so $U$ or $V$ is empty.

I'll add as a further note that I think the most useful characterization of connected spaces tends to be the second one. In particular whenever you need to show a connected space has some sort of property you show that the subset of $X$ having that property is both open and closed (also non-empty) and therefore all of $X$.

Edit: Fixed some errors.

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In (1) you want $U$ and $V$ to be disjoint. Your comment on proving that (3) implies (1) actually applies to proving that (1) implies (3). For (3) implies (1) one starts with the open partition $\{U,V\}$, defines $f$ by $f[U]=\{0\}$ and $f[V]=\{1\}$, uses (3) to say that it’s constant, and concludes that one of $U$ and $V$ must be empty. – Brian M. Scott Aug 30 '11 at 23:30
@Brian, you're completely right. My mistake. – JSchlather Aug 30 '11 at 23:40

I'm not sure if I understand your question, though it seems like you are asking for equivalent definitions for connectedness.

You can for example easily see that a space X is connected if and only if the only clopen (closed and open) sets are X and the empty set.

Also, it is always a good idea to take a look at the Wikipedia article on connectedness.

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Connectedness is not the same as path connectedness in general metric spaces... I think you should edit that part of your answer. – Sam Aug 30 '11 at 22:19
@Sam: Thanks! I should have checked my claim. – Fredrik Meyer Aug 30 '11 at 22:41

Assume there exists a continuous function f:X-->{0,1}, where {0,1} has the discrete topology. Then {0},{1} are both open in {0,1}, so that, by continuity of f:



are both open in $X$, and form a disconnection of $X$, i.e., $X=U\cup V$ , where $U,V$ are disjoint open sets in X.

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