# A topological space such that every subspace from it is connected.

Let $\tau$ be the topology in $\mathbb{R}^2$ whose open sets are of the form $$G_t = \{ (x,y)\in \mathbb{R}^2 \mid x>y+t\}, \; t\in \mathbb{R}$$ Is the union of two lines $r\cup s$ connected in $(\mathbb{R}^2,\tau)$?

What I did was consider $r\cup s$ as subspace of $\mathbb{R}^2$ and assume that therexist $A$ and $B$ such that $$r\cup s = A\cup B$$ and $A\cap B =\emptyset$. Then we can write $A$ and $B$ as $$A=(r\cup s)\cap G_{t_1},\, B= (r\cup s)\cap G_{t_2}$$ But since for every $t_1, t_2 \in \mathbb{R}$, $G_{t_1}\subset G_{t_2}$ or $G_{t_2}\subset G_{t_1}$, we have that $$A\subset B\; \text{or}\; B\subset A$$ In any case, $A\cap B\neq \emptyset$ and hence there is no separation of $r\cup s$, so $r\cup s$ is connected.

My question then is if this proof is okay, and also if it works too for any subspace of $(\mathbb{R}^2,\tau)$ since I haven't use at all anything about the two lines. Also

Are there other examples of topologies in $\mathbb{R}^2$ such that all subspaces are connected?

• The proof is okay, and it works in every space in which the open sets are nested. That’s a hereditary property, so it works for all subspaces of such spaces as well. – Brian M. Scott Aug 12 '16 at 17:15
• There is a flaw : We can write $A=(r\cup s)\cap G_{t1}\lor A=(r\cup s)\cap \mathbb R^2$ and similarly for $B$. But the conclusion $A\subset B\lor B\subset A$ will still hold. – DanielWainfleet Aug 13 '16 at 3:08

## 3 Answers

The spaces which have every subspace connected are essentially the total preorders EDIT: but see below.

In one direction, given a space $X$, define a relation $\le$ on $X$ given by $a\le b$ if every open set containing $b$, also contains $a$. This relation is clearly a preorder. If every subspace of $X$ is connected, then $\le$ is total: for every $a, b\in X$, either $a\le b$ or $b\le a$ (otherwise, $\{a, b\}$ would be a disconnected subspace of $X$).

Conversely, suppose $(L, \le)$ is a total preorder. Then we can turn $L$ into a topological space by taking as the opens those sets of the form $\{b: b\le a\}$ for $a\in L$. The preorder associated to this space is exactly $(L, \le)$. So this is an exact characterization.

This answer is actually incorrect in a subtle way, as Eric Wofsey points out below: the preorder assigned to a space does not characterize it! Specifically, we can have two spaces with isomorphic preorders, which are not homeomorphic. An easy example is: any two Hausdorff spaces with the same cardinality.

We can even come up with a hereditarily connected example! Let $X=\mathbb{N}\cup\{\infty\}$, and consider the following two topologies on $X$:

• $\tau=\{\{k, k+1, k+2, . . . \}\cup\{\infty\}: k\in\mathbb{N}\}\cup\{\emptyset\}$,

• $\sigma=\{\{k, k+1, k+2, . . . \}\cup\{\infty\}: k\in\mathbb{N}\}\cup\{\{\infty\}\}\cup\{\emptyset\}$.

It's easy to show that $\tau$ and $\sigma$ yield the same preorder.

However, this is a homeomorphism invariant for finite hereditarily connected spaces, or - even better - spaces whose associated preorder has finitely many equivalence classes. Roughly speaking, two hereditarily connected spaces with the same preorder differ only at the limit points in that preorder (although it's a bit messy to make this precise).

• Fine, but it might be better to describe them as the spaces whose every nonempty subspace is connected. – bof Aug 13 '16 at 2:29
• @bof The empty space is connected, so I don't think that's a problem. – Noah Schweber Aug 13 '16 at 2:31
• Depends on how you define connected. For some authors a space is connected if it has exactly one component. – bof Aug 13 '16 at 2:41
• @bof Isn't the emptyset its own unique connected component? It's a maximal subset which is connected in the seemingly-weaker sense. EDIT: Huh, nevermind, it's not a universal convention (math.stackexchange.com/questions/503442/…). Now I know! – Noah Schweber Aug 13 '16 at 2:49
• I used to think "the empty space is connected" was a universal convention, until recently when I learned better by reading some answer on his site, maybe the one you linked to. – bof Aug 13 '16 at 2:55

Noah's answer is not quite correct (edit: he has now edited it to comment on this). Given a topological space $X$, there is a preorder on $X$ defined by $a\leq b$ iff $b\in\overline{\{a\}}$, or equivalently if every open set containing $b$ contains $a$. This is called the specialization preorder. (There is no universal convention for which direction the order goes, so you will also see people define the specialization preorder as the opposite of how I have just defined it.)

As Noah showed, if every (nonempty) subset of $X$ is connected, then the specialization preorder must be total: if $a$ and $b$ are incomparable in the specialization preorder, then the subspace $\{a,b\}$ is disconnected. Conversely, if the specialization preorder of $X$ is total, then every (nonempty) subset of $X$ is connected: every two-point subset is connected, which means any two points are in the same connected component of any subspace that contains both of them.

So every (nonempty) subset of $X$ is connected iff the specialization preorder is total. And there are plenty of examples of topologies with total specialization preorder. Indeed, any preorder at all can be the specialization preorder of a topology: given a preorder $\leq$ on $X$, take the topology generated by the sets $\{y\in X:y\leq x\}$ for each $x\in X$.

However, it is not accurate to say that such topologies on a set $X$ are "the same" as total preorders on $X$, since different topologies can give the same preorder. For instance, let $X=\mathbb{R}$; then here are two different topologies whose specialization preorder is the usual order on $\mathbb{R}$. The first topology is generated by the sets $(-\infty,x]$ for each $x\in\mathbb{R}$. The second topology is generated by the sets $(-\infty,x)$ for each $x\in\mathbb{R}$.

The indiscrete topology $\{\mathbb R ^2,\varnothing\}$ also works. Or for a $T_1$ example, let $X:=\{\langle n,0\rangle:n\in\omega\}$ and take a basis of sets $U\subseteq \mathbb R ^2$ such that $U=V\cup (X\setminus F)$ for some open $V$ in $\mathbb R ^2$ and $F\subseteq X$ finite. You cannot get better than $T_1$ (if the space is Hausdorff, then any set of size $2$ is not connected).

By the way, the beginning space $\mathbb R ^2$ does not matter for these examples.