Is there a topological space $(C,\tau_C)$ and two points $c_0\neq c_1\in C$ such that the following holds?

A space $(X,\tau)$ is connected if and only if for all $x,y\in X$ there is a continuous map $f:C\to X$ such that $f(c_0) = x$ and $f(c_1) = y$.

Is there also a Hausdorff space satisfying the above?


4 Answers 4


No such space $C$ can exist.

We will derive a contradiction from the assumption that $C,c_0,c_1$ as desired exists.

Let κ be any cardinal greater than $|C|$. View $\kappa$ as an ordinal. For each β in κ add a copy of the unit interval between β and β+1, and add a point ∞ at the end. The resulting "Very Long Line" $L_\kappa$ is dense and complete, hence connected as a topological space.

As $L_\kappa$ is connected, there must be a continuous map $f$ from $C$ into $L_\kappa$ whose image $L'$ contains $0$ and $\infty$. Let $b\notin L'$. Then the map $h$ that sends everything below $b$ to $0$ and everything above $b$ to $1$ is continuous from $L'$ to the discrete space $\{0,1\}$.

So the map $h\circ f:C\to \{0,1\}$ witnesses that $\{0,1\}$ is connected, a contradiction.

(This is just a variant of Helene Sigloch's earlier argument.)


First, there can't be a path from $c_0$ to $c_1$, else a continuous map would give a path from $x$ to $y$. By the same argument, $C$ is not allowed to have finitely many or countably many path components. If one can build something like the topologist's sine curve on a very long line of arbitrarily large cardinality, the number of path components of $C$ has to be larger or equal than any cardinal number. There is no largest cardinal number. Hence, the set of points of $C$ is no set.

Thus: If there exist long lines in every cardinality, the answer to your question is no.

Edit: OK, if $x$ and $y$ are uncountably far apart, there is no path from $x$ to $y$. Call the set of points between $x$ and $y$ the "long path" from $x$ to $y$. It contains an arbitrarily large number of connected non-path-connected, pairwise disjoint sub"paths". Say the number is $\kappa '$. The connected components of their preimages under $f$ also have to be non-path-connected and pairwise disjoint. Hence, the universal space $C$ that we are looking for needs to have at least $\kappa '$ path components, hence at least $\kappa '$ points. If the "very long line" exists for every cardinality, the universal space $(C, c_0,c_1)$ can't exist.

Edit: Goldstern's argument is much simpler than this. Just for completeness: Very long lines exist (see Goldstern's comment) and so the answer is definitely No.

Edit: The concept of homotopy theory with respect to a big interval is actually worked with. I just stumbled over a preprint by Penrod in the field and the foundations seem to be laid by Cannon-Gonner [J. Cannon and G. Conner, The big fundamental group, big Hawaiian earrings, and the big free groups, Topology Appl. 106 (2000)].

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    $\begingroup$ I'm not sure I follow your argument. I think it has to be shown that for any space $(C,\tau_C)$ and points $c_0,c_1\in C$ there is a space $(X,τ)$ that is connected, but not "$(C,c_0,c_1)$-connected", or vice versa. $\endgroup$ Mar 24, 2015 at 9:47
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    $\begingroup$ Very long lines exist. Let $\kappa$ be any cardinal (viewed as an ordinal). For each $\beta$ in $\kappa$ add a copy of the unit interval between $\beta$ and $\beta+1$, plus a point $\infty$. The resulting linear order is dense and complete. But if $C$ has cardinality smaller than $\kappa$, then any continuous image of $C$ that contains $0$ and $\infty$ will not be onto, hence not connected. $\endgroup$
    – Goldstern
    Mar 24, 2015 at 12:26
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    $\begingroup$ @Goldstern Right. For every topological space there exists a connected topological space that it doesn't surject to. This is a nice simple argument. You should make it an answer. $\endgroup$ Mar 24, 2015 at 12:48
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    $\begingroup$ Why does the existence of many path components of $X$ imply as many of $C$ when $f$ is not required to be surjective? $\endgroup$
    – LSpice
    Mar 24, 2015 at 12:52
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    $\begingroup$ $f$ is required to be continuous and to hit both $x$ and $y$. That means it has to hit the whole line segment between $x$ and $y$. $\endgroup$ Mar 24, 2015 at 12:57

I claim that for each cardinal $\lambda$, there is a connected space $C$ and $c_{0},c_{1}\in C$ such that whenever $|X|<\lambda$, then $X$ is connected if and only if for all $x,y\in X$ there is some continuous map $f:C\rightarrow X$ with $f(c_{0})=x$ and $f(c_{1})=y$.

Let $I$ be an index set and let $(X_{i},x_{i},y_{i})_{i\in I}$ be an enumeration of all the connected spaces with two points of cardinality less than $\lambda$ up to an isomorphism preserving the $x_{i}$ and $y_{i}$. Let $C=\prod_{i\in I}X_{i}$, let $c_{0}=(x_{i})_{i\in I}$, and let $c_{1}=(y_{i})_{i\in I}$. Then $C$ is connected. Therefore if $X$ is a space and for each $x,y\in X$ there is some $f:C\rightarrow X$ with $f(c_{0})=x,f(c_{1})=y$, then $X$ is connected. On the other hand, if $X$ is connected, $|X|<\lambda$ and $x,y\in X$, then $(X,x,y)\simeq(X_{i},x_{i},y_{i})$ for some $i\in I$. However, the projection onto the $i$-th coordinate $\pi_{i}:C\rightarrow X_{i}$ is a continuous mapping with $\pi_{i}(c_{0})=x_{i},\pi_{i}(c_{1})=y_{i}$.

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    $\begingroup$ Cool! This confirmatively answers the implicit follow-up after the initial negative answer. $\endgroup$
    – jmc
    Mar 24, 2015 at 15:32
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    $\begingroup$ Very nice, Joseph! $\endgroup$ Mar 24, 2015 at 16:04
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    $\begingroup$ This $\ C\ $ works only for a given in advance family of connected spaces, and there is no any sharp upper bound on the size of $\ C.\ $ Thus I fail to see this construction as cool or very nice. On the contrary, it is routine, and it doesn't buy much. $\endgroup$ Mar 28, 2015 at 6:41
  • $\begingroup$ A more interesting follow-up question would be: Given $\kappa$, is there a not too large space (say: of weight $\kappa$, or perhaps cardinality or density $\kappa$, or perhaps we might allow $\kappa^+$ or $2^\kappa$?) which decides connectedness (in the way suggested by Dominic) for all small spaces (or weight or cardinality or density $\le \kappa$)? $\endgroup$
    – Goldstern
    Mar 28, 2015 at 21:26
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    $\begingroup$ While this construction may be too trivial to have much mathematical content, I think it may well have some metamathematical content, by helping to explain why many results concerning path-connectedness seem to "automatically" have analogues for topological connectedness (or vice versa). For instance, from this construction, one can adapt the proof of (say) the fact that the product of two path-connected spaces is path-connected, to conclude the same for topologically connected spaces (and note that the cardinality restriction is not a probem here). $\endgroup$
    – Terry Tao
    May 9, 2015 at 19:30

I hope that my example below is as elegant as the continuous long line provided by Goldstern above, while my example is less expected. Also, while long line is simpler in itself, the proof is simpler in my case. Finally, perhaps logicians will find some advantages (I'll do a little of it--I am not confident to do it well).

Let $\ A\ $ be an arbitrary set. The following ordered triple $\ (\mathbf S_A\ \mathbf 0\ \mathbf 1)\ $, where $\ \mathbf S_A:=(S_A\ T_A)\ $ is a topological space--call it a skeleton, and $\ \mathbf {0\ 1}\in S_A,\, $ is to be defined below, while first (ahead of time) let's formulate

THEOREM   For every connected subset $\ X\subseteq S_A,\ $ such that $\ \mathbf {0\ 1}\in X,\ $ the inequality of cardinalities $\ |X|\ge|A|\ $ holds.

This instantly gives a simple negative answer to the question of this thread posed by Dominic.


  • $\ S_A\ :=\ \{(x_a)_{a\in A}\in[0;1]^A\ :\ \forall_{a\ b\in A} [x_a\ x_b\in(0;1)\ \Rightarrow\ a=b]\ \}$
  • $\ \mathbf 0\ :=\ (0)_{a\in A}\ $ and $\ \mathbf 1\ :=\ (1)_{a\in A}$
  • $\ T_A\ $ is the topology in $\ S_A\ $ induced by the Tikhonov toplogy in cube $\ [0;1]^A$

PROOF (of the theorem)   The connected component of $\ \mathbf 0\ $ in $\ S_A\ $ is dense in $\ S_A,\ $ which means that its closure, i.e. space $\ S_A\ $ itself, is connected too. Next, let:

$$H^a\ :=\ \{x\in[0;1]^A\ :\ x_a=\frac 12$$

Let $\ X\subseteq S_A\ $ be a connected subset such that $\ \mathbf {0\ 1}\in X.\ $ Then $\ p_a(X)=[0;1],\ $ hence $\ H\,^a\cap X\ne \emptyset.\ $ Thus

$$ |X|\ \ge\ \left|\left\{H^a\ :\ a\in A\ \right\}\right|\ =\ |A|$$

Indeed, sets $\ H^a\ $ are disjoint (hence different).   END of Proof

            G E N E R A L I Z A T I O N

We may replace the topological interval $\ [0;1],\ $ and its three points $\ 0\ \frac 12\ 1,\ $ by an arbitrary connected space $\ S\ $, and its three points $\ a\ h\ b,\ $ such that $\ h\ $ separates $\ a\ b\ $ (meaning that there are open sets $\ G\ $ and $\ H:=(S\setminus\{h\})\setminus G\ $ of $\ S\ $ such that $\ a\in G\ $ and $\ b\in H$. Etc. The theorem still holds.

Logical considerations

I am not using ordinal numbers. My construction is free of any complications, especially when $\ S\ $ of the generalization is a proper 3-point space $\ \{a\ h\ b\}.\ $ Thus I am worried only about axioms like the axiom of choice or continuum hypothesis, and similar, about their relation to the cartesian product, and to the ordinary $\ [0;1]\ $ of my main example.

EXTRA. Compactness. Another connectedness proof.

Space $\ \mathbf S_A\ $ is compact, it is a closed subset of the Tikhonov cube $\ [0;1]^A:\ $ indeed, let $\ x\in [0;1]^A\setminus S_A.\ $ Then there exist two different indices $\ a\ b\ \in\ A\ $ such that $\ (x_a\ x_b)\in (0;1)^{\{a\ b\}}.\ $ Thus the inverse image of this open square under the canonical projection $\ p_{a\ b} : [0;1]^A\rightarrow(0;1)^{\{a\ b\}}\ $ is disjoint from $\ S_A\ $ (one could say that $\ [0;1]^A\setminus S_A\ $ is open because it is a union of the inverse images of the open squares). Thus indeed $\ S_A\ $ is compact.

Now $\ \mathbf S_A\ $ is connected because it is an inverse limit of spaces $\ \mathbf S_B\ $ for all finite $\ B\subseteq A,\ $ under the canonical projections. (One could also use som other similar arguments). This inverse limit nature of $\ \mathbf S_A\ $ shows its covering 1-dimensionality:

$$\dim \mathbf(S_A)\ =\ 1$$


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