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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.

DEFINITION

• $\ S_A\ :=\ \{(x_a)_{a\in A}\in[0;1]^A\ :\ \forall_{a\ b\in A} [x_a\ x_b\in(0;1)\ \Rightarrow\ a=b]\ \}$
• $\ \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\ 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:

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

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.

DEFINITION

• $\ S_A\ :=\ \{(x_a)_{a\in A}\in[0;1]^A\ :\ \forall_{a\ b\in A} [x_a\ x_b\in(0;1)\ \Rightarrow\ a=b]\ \}$
• $\ \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:

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

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 formulated

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.

DEFINITION

• $\ S_A\ :=\ \{(x_a)_{a\in A}\in[0;1]^A\ :\ \forall_{a\ b\in A} [x_a\ x_b\in(0;1)\ \Rightarrow\ a=b]\ \}$
• $\ \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\ 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:

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

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.

DEFINITION

• $\ S_A\ :=\ \{(x_a)_{a\in A}\in[0;1]^A\ :\ \forall_{a\ b\in A} [x_a\ x_b\in(0;1)\ \Rightarrow\ a=b]\ \}$
• $\ \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\ 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:

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

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 formulated

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.

DEFINITION

• $\ S_A\ :=\ \{(x_a)_{a\in A}\in[0;1]^A\ :\ \forall_{a\ b\in A} [x_a\ x_b\in(0;1)\ \Rightarrow\ a=b]\ \}$
• $\ \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\ 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).

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 formulated

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.

DEFINITION

• $\ S_A\ :=\ \{(x_a)_{a\in A}\in[0;1]^A\ :\ \forall_{a\ b\in A} [x_a\ x_b\in(0;1)\ \Rightarrow\ a=b]\ \}$
• $\ \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\ 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:

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