I'm learning about connected topological spaces, and i came up with a related question i'm not able to answer. To make things simpler, i'm going to give some definitions. Let $X$ be a topological space. For a positive integer $n \in \mathbb{N}$, $n \geq 2$ we say that $X$ is $n$-separated if there exist non empty and pairwise disjoint open subsets $U_1, …, U_n \subset X$ such that $X = U_1 \cup \cdot \cdot \cdot \cup U_n$, furthermore, we say that $X$ is countably separated if there exists a sequence $\{ U_n \}_{n=1}^\infty$ of non empty pairwise disjoint open subsets of $X$ such that $X = \cup_{i=1}^\infty U_i$. It's obvious that if a topological space is countably separated then it is $n$-separated for all $n \geq 2$. Now i'm trying to understand if the converse is true. The only thing i was able to notice is that if a topological space is $n$-separated for all $n \geq 2$, then it has infinitely many connected components. Indeed, suppose by contradiction that $C_1, …, C_m \subseteq X$ are all the connected components of $X$, for some $m \in \mathbb{N}$, $m \geq 1$. By definition there are non empty and pairwise disjoint open subsets $U_1, …, U_{m+1} \subseteq X$ such that $X = \cup_{i=1}^{m+1} U_i$. Notice that each $C_i$ is contained in one of the $U_j 's$ (Suppose by contradiction that $C_i$ intersects more than one of the $U_j$'s. Without loss of generality we can reorder the $U_j$'s and suppose that there is an $1 < r \leq m+1$ such that $C_i$ intersects $U_1, …, U_r$ and doesn't intersect $U_j$ for $j > r$. This means that $C_i = C_i \cap (U_1 \cup U_2 \cup \cdot \cdot \cdot \cup U_r) = (C_i \cap U_1) \cup (C_i \cap (U_2 \cup \cdot \cdot \cdot \cup U_r))$, in contradiction with the fact that $C_i$ is connected) and therefore, because $m + 1 > m$, this implies that at least one of the $U_i$'s is empty. Contradiction.
Weird generalization of disconnected topological spaces
connectednessgeneral-topology
Related Solutions
Here is a construction in the $2$-sphere $S^2$, equipped with any reasonable metric. By removing one point it becomes homeomorphic to the plane, so it gives an example in $\mathbb{R}^2$. (You have to be a little careful which point to remove, but it is not that hard to figure out that there exists one that works. Alternatively, equip $\mathbb{R}^2$ with a bounded metric and run the same construction.) The construction is similar to the standard "Lakes of Wada" construction in spirit.
Let $U_1^1$ be a simple path which is $1$-dense in $S^2$, i.e., such that every point on the sphere has distance $\le 1$ to a point on $U_1^1$. Now let $U_2^1$ be a simple path (i.e., a homeomorphic image of $[0,1]$) in $S^2 \setminus U_1^1$ which is $1$-dense in $S^2$. Proceed to get disjoint $1$-dense simple paths $U_1^1,\ldots U_n^1$. Now extend $U_1^1$ to obtain a $1/2$-dense simple path $U_1^2$ in $S^2 \setminus (\bigcup_k U_k^1)$. Inductively construct a sequence of mutually disjoint simple paths $U_1^2,\ldots U_n^2$ which are $1/2$-dense extensions of $U_1^1,\ldots,U_n^1$. Now keep extending those inductively to get mutually disjoint paths $U_1^m,\ldots U_n^m$ which are $1/m$-dense in $S^2$. This construction is possible because at any step the complement of the already constructed paths is connected, since it is the complement in $S^2$ of a finite set of disjoint homeomorphic images of $[0,1]$.
Now let $U_k^\infty = \bigcup_m U_k^m$ for $k=1,\ldots,n$. This is a collection of mutually disjoint open paths (continuous images of $[0,1)$ or $(0,1)$, depending on how exactly the extensions are chosen), each of them dense in the plane. Their union is not necessarily all of $\mathbb{R}^2$, so let $T=S^2 \setminus \bigcup_k U_k^\infty$, and let $U_1 = U_1^\infty \cup T$ and $U_k = U_k^\infty$ for $k\ge 2$. Then $S^2 = \bigcup_k U_k$ is a disjoint partition, and since $U_2,\ldots,U_n$ are continuous images of an interval, they are connected, even path-connected. The set $U_1$ is not necessarily path-connected, so in order to show connectedness assume that $U_1 = A \cup B$ with relatively open disjoint sets $A$ and $B$. Since $U_1^\infty$ is path-connected, it has to be contained in either $A$ or $B$. We may assume $U_1^\infty \subseteq A$. Assume $t \in T \cap B$. Since $U_1^\infty$ is dense and $B$ is relatively open, there has to exist $u \in U_1^\infty \cap B$. However, this contradicts $A \cap B = \emptyset$.
The last argument is probably some standard topology result, that if $U$ is connected, and $V\supseteq U$ is contained in the closure of $U$, then $V$ is connected. The crucial point is to find disjoint connected dense subsets in the first place.
This construction does not guarantee that $U_1$ is path-connected, and I am not sure whether the similar question about a path-connected partition has a positive answer.
There are two possibilities:
- $A\cap B=\emptyset$: then $A\cap B$ is connected.
- $A\cap B\neq\emptyset$: then $A\cup B$ is connected, because if $f\colon A\cup B\longrightarrow\{0,1\}$ (with $\{0,1\}$ endowed with the discrete topology) is continuous then, if $p\in A\cap B$, $f(A)=\bigl\{f(p)\bigr\}=f(B)$, and therefore $f$ is constant.
Best Answer
Nice question - indeed, the converse need not hold in general!
Consider the topology on $\mathbb{N}$ (which for me contains $0$) generated by:
all singletons $\{n\}$ with $n>0$, and
all cofinite sets.
Basically, a set is open in this topology iff it either doesn't contain $0$ or it is cofinite. (Incidentally, this is the one-point compactification of the discrete topology on $\mathbb{N}_{>0}$.)
This is clearly $n$-disconnected for each $n$, but there is no infinite collection of disjoint open sets whose union contains $0$ at all.