Is it true for arbitrary topological spaces?

Question

In Rudin's book "Principles of Mathematical Analysis" there is a definition for sets to be disconnected: $A\subset X$ is disconnected if there exists an open cover $\{U,V\} $ of $A$ such that $A\subset U\cup V$ and $\bar{U}\cap V=U\cap\bar{V} =\emptyset $, under the hypothesis that $X$ is a metric space. Does this holds for an arbitrary topological space, i.e. for $X$ to be a topological space, a set $A\subset X$ is disconnected if and only if there exists an open cover $\{U,V\} $ of $A$ in $X$ such that $\bar{U}\cap V=V\cap\bar{U} =\emptyset $ ,$A\cap U\neq\emptyset $ and $A\cap V\neq\emptyset $? If it is wrong, is there an counterexample?

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I have made two main mistakes. What I actually want is a conclusion that from the disconnectedness of a subspace $F$ of a given topological space $X$ I can find two disjoint open sets in $X$ such that they forms an open cover of $F$ and $F$ intersects with both sets, but it is wrong because the property is equivalent to the complete normality of a topological space(in Munkres' book "Topology" there is an exercise talking about this, see section 32), so it cannot be true for an arbitrary topological space. Another mistake I make is that I have misread the definition of connectedness in Rudin's book.

Answer 1

Counter example. A = (0,1) = U, V = (2,3).
As U and V are open, it is prefered to simply require
them to be disjoint instead of separated.
Did not Rudin require a point of A be in U and
another point of A to be in V?

Answer 2

Rudin's actual definition (for the reals really) (on page 42):

two sets $A$ and $B$ are called separated iff $A \cap \overline{B} = \overline{A} \cap B = \emptyset$. A set $E \subseteq \mathbb{R}$ is called connected when it cannot be written as the union of two non-empty separated sets.

This definition is correct in all spaces, although the usual topology definition is intrinsic in nature, not "connected subset of $X$" but connected space $X$:

A space $X$ is connected if we cannot write it as a union of two disjoint non-empty open subsets (or equivalently, closed subsets). If $A \subseteq X$ is a subset of $X$ we $A$ is called connected when $A$ in its subspace topology wrt $X$ is a connected space.

You can try to show that this coincides with the definition that Rudin gave above. (I might fill it in later, time permitting)

Answer 3

The equivalence is easy to prove: if $U$ and $V$ are open sets in any topological space with $U\cap V=\emptyset$ then $U \subset V^{c}$ and $V^{c}$ is closed so $\overset {-} U \subset V^{c}$ so $\overset {-} U \cap V=\emptyset$.