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.

  • 1
    $\begingroup$ No. Hint: Use $X$ with the particular point topology. $\endgroup$ Commented May 26 at 23:38
  • $\begingroup$ My bad. By "topological space" I meant "Hausdorff topological space". I'm very sorry for the misunderstanding. $\endgroup$
    – Peluso
    Commented May 27 at 0:01
  • 3
    $\begingroup$ You should think of any other forgotten assumptions.... $\endgroup$ Commented May 27 at 0:08
  • $\begingroup$ Understood. Again, I apologize. $\endgroup$
    – Peluso
    Commented May 27 at 0:40

1 Answer 1


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.

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

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