# Is this an equivalence of connectedness?

Let $$X$$ be a Hausdorff topological space and $$A\subseteq X$$. Suppose that for every $$B\subseteq X$$, $$A\cap Bd(B)\neq \emptyset$$ (i.e., $$A$$ has non-empty intersection with the boundary of $$B$$) whenever $$A\cap B \neq \emptyset \neq A\cap (X\setminus B)$$. Must $$A$$ be connected? Clearly, the other implication always holds, but I have not been able to determine if it is actually an equivalence of connectedness.

• No. Hint: Use $X$ with the particular point topology. Commented May 26 at 23:38
• My bad. By "topological space" I meant "Hausdorff topological space". I'm very sorry for the misunderstanding. Commented May 27 at 0:01
• You should think of any other forgotten assumptions.... Commented May 27 at 0:08
• Understood. Again, I apologize. Commented May 27 at 0:40

Here is another example, it is Hausdorff but not regular (I am assuming that you did not forget to assume regularity!). Let $$X$$ be the rational upper half-plane with the irrational slope topology. It is an example of a countable, connected, Hausdorff (but not regular) space. Let $$A\subset X$$ be the subset consisting of points with nonzero second coordinate. When equipped with the subspace topology, $$A$$ is discrete, hence, totally disconnected. Below, I will prove that $$A$$ satisfies your condition.

Given $$z\in U=\mathbb R\times (0,\infty)$$ (the upper half-plane), we have two lines $$L_\pm$$ with slopes $$\pm \theta$$ through $$z$$ (where $$\theta$$ is the fixed irrational number used to define the irrational slope topology). Let $$z_\pm$$ denote the intersection points of $$L_\pm$$ with the $$x$$-axis; if $$z\in A$$, these are, of course, irrational numbers. We obtain the correspondence $$P: U\to \mathbb R$$ sending $$z$$ to $$\{z_-, z_+\}$$.

We now fix a subset $$B\subset X$$ such that $$\partial_X B\cap A= \emptyset$$. Set $$B_1:=B\cap A$$, $$B_2=B\cap \mathbb Q\times \{0\}$$. I will assume that $$B_1\ne \emptyset$$. Then $$B=B_1\sqcup B_2$$. I will be identifying $$\mathbb Q\times \{0\}$$ and $$\mathbb Q$$.

The definition of the irrational slope topology implies:

Lemma 1. Suppose that $$C\subset \mathbb Q$$ is such that for $$a\in A$$, $$P(z)$$ has nonempty intersection with the closure of $$C$$ in $$\mathbb R$$. Then $$a\in cl_X(C\times \{0\})$$.

Corollary 1. $$P(B_1)$$ does not intersects the accumulation set of the complement of $$B$$ in $$\mathbb Q$$. In other words, $$P(B_1)$$ is contained in the interior of the closure of $$B_2$$ in $$\mathbb R$$.

Since $$B_1$$ is nonempty, there exists an open interval $$I\subset \mathbb R$$ such that $$cl_{\mathbb R}(B_2)$$ contains $$I$$. Consider $$P^{-1}(I)$$: It is a union of two open strips in $$U$$ with the slopes $$\pm \theta$$. Rational points are dense in these strips. Therefore, $$P(P^{-1}(I)\cap A)$$ is dense in $$\mathbb R$$.

Lemma 2. $$P^{-1}(I)\cap A\subset B_1$$.

Proof. Since $$I\subset cl(B_2)$$, by Lemma 1 each $$a\in P^{-1}(I)\cap A$$ is a limit point of $$B_2$$ in $$X$$. If $$a\notin B_1$$ then $$a\in \partial_X B_1\cap A$$, which contradicts the assumption on $$B$$. qed

By combining Lemma 2 with the preceding observations, we obtain Corollary 2:

Corollary 2. $$P(B_1)$$ is dense in $$\mathbb R$$.

Corollary 3. $$B_2$$ is dense in $$\mathbb R$$.

Proof. By Corollary 1, $$P(B_1)$$ is contained in the closure of $$B_2$$ in $$\mathbb R$$. Denseness of $$P(B_1)$$ in $$\mathbb R$$ implies that $$B_2$$ is dense in $$\mathbb R$$ as well.

Lemma 3. $$A\subset B$$.

Proof. Suppose there is $$a\in A\setminus B$$. By Corollary 3, $$P(a)$$ is contained in the closure of $$B_2$$ in $$\mathbb R$$. Then, by Lemma 1, $$a\in cl_X(B_2)$$. But then $$a\in \partial_X B_2$$. This contradicts the assumption on $$B$$. qed

Thus, we proved:

Theorem. There is no subset $$B\subset X$$ such that $$B\cap A\ne \emptyset\ne A\setminus B$$ and $$\partial_X B\cap A=\emptyset$$.

In particular, your definition is not equivalent to connectedness for subsets of Hausdorff spaces.

• Indeed, I did not forget regularity! Thank you very much for your patience and for your example. Commented May 29 at 1:14