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Defintions:

Let $X$ be a topological space.

1) A connected space $Y$ is a minimal connected ambient (m.c.a for short) space for $X$ if there exists an embedding $i:X\mapsto Y$, and for every connected space $Y'$ into which $X$ can be embedded, there exists an embedding $j:Y\mapsto Y'$.

2) A connected space $Y$ is a smallest connected ambient (s.c.a for short) space for $X$ if there exists an embedding $i:X\mapsto Y$, and for every connected space $Y'$ into which $X$ can be embedded, and for any embedding $j:X\mapsto Y'$ it holds that $|Y-i(X)|\leq |Y'-j(X)|$. (That is we added the least amount of points to $X$ to create $Y$).

Remarks:

1) If $X$ is connected then clearly $X$ itslef is both a minimal and smallest connected ambient space.

2) As noted by John, every topological space $X$ can be embedded into it's cone $CX$, which is connected.

This raises some quetions:

Assume $X$ is not connected.

1) Is it always possible to embedd $X$ in a smallest (s.c.a) connected ambient space?

Udate: The answer is yes. (According to the construction given by Kaj Hansen, which is the open extension). Since we assume $X$ is not connected, any connected space in which $X$ embedds must contain points that are not in the image of $X$. So it must contain at least one more point. Now this trivial lower bound is achieved by the wonderful construction of Hansen).

2) Is it always possible to embedd $X$ in a minimal (m.c.a) connected ambient space? I suspect the answer is nt, and that for most spaces $X$ an there will not be an m.c.a.

3) When the m.c.a (s.c.a) exists, are they unique? (=are they homeomorphic? or even hoemomorphic by a homeomorphism which takes the copy of $X$ in one of them into the copy in the other?)

4) If $Y$ is an m.c.a for $X$ is it also an s.c.a for it? (What about the other direction?)

5) What happens when we add additional structue? (in particular order structure).

Let $(X,<_X)$ be an ordered topological space.

Say $(Y,<_Y)$ is a minimal ordered connected ambient (m.o.c.a for short) space for $X$ if there exists an order-preserving embedding $i:(X,<_X)\mapsto (Y,<_Y)$, and for every ordered connected space $(Y',<_Y')$ into which $(X,<)$ can be embedded , there exists an (order-preserving) embedding $j:(Y,<_Y)\mapsto (Y',<_Y')$.

We define similarly (see def2 above) a smallest ordered connected ambient (s.o.c.a for short) space for $X$

Now we can ask: are there always s.o.c.a, m.o.c.a?

Example:

$\mathbb{Q}$ is totally disconnected. (In the order topology). However it can be embedded in $\mathbb{R}$ which is connected. (This is because the order topology in $\mathbb{Q}$ and the subspace topology on it inherited from $\mathbb{R}$ coincide).

$\mathbb{R}$ is both an m.o.c.a and an s.o.c.a for $\mathbb{Q}$. This follows from the fact that every ordered topological space which is connected must satisfy the completeness axiom about existence of suprema, and hence if it contains $(\mathbb{Q},<_\mathbb{Q})$ it must contain a copy of $(\mathbb{R},<_\mathbb{R})$.

Note that if we do not require the ambient space to be ordered, there is a smaller suitable ambient space (the open extension). In other words the s.c.a and s.o.c.a are differnt in this case.

Asaf Shachar
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    Every topological space $X$ can be embedded into it's cone $CX$, which is connected. –  May 28 '15 at 07:54
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    It is not so clear how 'minimal" can be probably defined. –  May 28 '15 at 08:00
  • @John: You are right in both comments. I will edit the question accordingly. – Asaf Shachar May 28 '15 at 08:04
  • I have a deja vu. – Asaf Karagila May 28 '15 at 10:03
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    For the minimal ‘connectification’ do you require that $Y$ embed in $Y'$ in a way that preserves the embedding of $X$? That is, if $i$ embeds $X$ in $Y$, and $i'$ embeds $X$ into $Y'$, do you require that there be an embedding $j:Y\to Y'$ such that $j\circ i=i'$? – Brian M. Scott Jun 01 '15 at 21:16
  • @Brian: This is a very good question. I thought about it and decided not to require this property in order to make the definition less rigid. Of course the composition $j \circ i:X\rightarrow Y'$ will always be some embedding (since a composition of topological embeddings is a topological embedding). However I am not sure about how the answer changes if we do add this requirement. In a sense it looks like a natural condition to impose. In summary I would be happy to see an answer to each of the versions. – Asaf Shachar Jun 02 '15 at 09:20
  • In the particular case of $\mathbb{Q}, \mathbb{R}$, if we only require that the (m.o.c.a) for $\mathbb{Q}$ to be an ordered set (without a field structure), or even if it is a field we require only order-embedding of $\mathbb{Q}$ in it, then we can take another copy of $\mathbb{R}$ and embedd $\mathbb{Q}$ in it with an additive shift so the natural map (the field-isomorphism) between these two copies of $\mathbb{R}$ won't preserve the embeddings of $\mathbb{Q}$ . However, since we work only in the category of ordered sets and not ordred fields, – Asaf Shachar Jun 02 '15 at 09:38
  • we do have an order-embedding between these two copies of $\mathbb{R}$, that preserves the embeddings of $\mathbb{Q}$. (Essentially we map one copy of $\mathbb{Q}$ onto the other one (via order-embedding) and extend this map by taking supremum of a certain set of rationals to the supremum of the corrseponding rational set in the codomain). If we require in addition that the (m.o.c.a) will be a field and the embeding to be a field-isomorphism, then preservation of the embedding $\mathbb{Q}$ must hold. (Isomorphism between fields of characteristic $0$ preserve their copies of $\mathbb{Q}$. – Asaf Shachar Jun 02 '15 at 09:41
  • Asaf: Let $X$ be the two-point order. You can embed it as the endpoints of $Y=[0,1]$. You can also embed it as the endpoints of the one-point compactification of the closed long ray; call this $Y'$ and let $p$ and $q$ be the left and right endpoints, respectively. (Thus, $q$ is the point added in compactification.) Then $Y$ and $Y'$ are both connectifications of $X$, and $Y$ embeds in $Y'$, but there is no embedding of $Y$ in $Y'$ that takes $0$ to $p$ and $1$ to $q$. You could replace $Y'$ by the lexicographically ordered square. – Brian M. Scott Jun 02 '15 at 15:49

2 Answers2

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Here is an idea that I think might be worth your consideration, though it does not take an ordering into account.

Consider a set $X$ endowed with the discrete topology: a"worst case scenario" as far as connectedness is concerned. Add a single point to $X$—call it $\delta$—and define the following topology on $X \cup \{\delta \}$:

  • If a subset $U \subset X \cup \{ \delta \}$ does not contain $\delta$, then it is open, as it would be in $X$ under the discrete topology.

  • If a subset $U \subset X \cup \{ \delta \}$ contains $\delta$, then $U$ is open $\iff U = X \cup \{ \delta \}$.

Notice that we have preserved the original topology on $X$, so there is a natural embedding $X \hookrightarrow X \cup \{ \delta \}$. Moreover, the only open set containing $\delta$ is the whole space! Thus, $X \cup \{ \delta \}$ cannot be expressed as the union of two nonempty, disjoint open sets. In other words, $X \cup \{ \delta \}$ is connected. Note also that this works regardless of the original topology on $X$: replace the first bullet with "If $\delta \notin U$, then $U$ is open $\iff$ it is open in $X$" .

So not only is this a one-point compactification of $X$; it's also a one-point connectification.

P.S. this general idea has been very kind to me in the past, so it might be worth bearing in mind. See here, for instance, where it can be used to show that the intersection of two compact sets need not be compact.

Kaj Hansen
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    I was about to write that term "one point connectification" in the comment..... –  May 28 '15 at 08:21
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    In the terminology of Steen and Seebach's Counterexamples in Topology, this is the open extension of $X$. They also note that it is connected, as reflected at https://proofwiki.org/wiki/Open_Extension_Space_is_Connected – Chris Culter May 28 '15 at 08:28
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    Very cool @ChrisCulter ! I was unaware that this example had been recorded elsewhere, but it doesn't surprise me in the least. Thanks for the reference. As a side note, the general idea here has been very kind to me in the past. See my post here, for example: http://math.stackexchange.com/questions/1276721/give-examples-of-compact-spaces-a-b-such-that-a-cap-b-is-not-compact/1276731 – Kaj Hansen May 28 '15 at 08:32
  • To take the ordering into account, one can define $x \le \delta$ for all $x\in X$. –  May 28 '15 at 08:33
  • By the way, it isn't quite right to call this the one-point compactification of $X$. In a one-point compactification, the added point typically does have neighborhoods that aren't the whole space. It happens that the one-point compactification of $\mathbb Q$ is also connected, so it's another good example. – Chris Culter May 28 '15 at 08:34
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    That crossed my mind @John, but we should be careful to note that the topology that's been defined will be strictly coarser than the order topology. – Kaj Hansen May 28 '15 at 08:35
  • ...oh actually, you said a one-point compactification, not the one-point compactification. Never mind then. :) – Chris Culter May 28 '15 at 08:36
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    Indeed @ChrisCulter. Of course something like the Alexandroff compactification would be much nicer if we were trying to preserve other topological properties. But connectedness was the priority here. :P – Kaj Hansen May 28 '15 at 08:38
  • @John: I think your suggestion to define the order $x\leq \delta$ does not work. When you do this, you get additional open sets of the form ${x>x_0}$ where $x_0 \in X$. In particular the new ordered space will be connected if and only if the original $X$ was connected. That is because as I said in the exmaple gave (end of the question) an ordered space which is connected must satisfy the completness axiom. So if the original set did not satisfy it then also $X \cup {\delta} $ won't satisfy it. Think about $\mathbb{Q}$ again. – Asaf Shachar May 28 '15 at 09:31
  • The new space will be $\mathbb{Q} \cup {\delta} $ and we can decompose it as follows: $(-\infty , \sqrt 2) \cup (\sqrt 2 , \infty)$. In this case it will hold that $(\sqrt 2 , \infty)= (\sqrt 2 , \delta]$ but it does not matter. To conclude: There is no way to get away from "adding all these suprema" (that is all the irrationals) in order to get a connected space. – Asaf Shachar May 28 '15 at 09:32
  • In response to all your comments so far, I have extended and changed the question thoroughly. – Asaf Shachar May 28 '15 at 09:35
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If the totally disconnected space is second countable then, since it is normal, it can be embedded in R^(infinity), a metrisable connected space.

Adelafif
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