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I am struggling with this old problem, which is also posted herehere:

Let $X$ satisfy countable chain condition(abbreviated as CCC) and $X$ has a regular $G_\delta$-diagonal. Then the cardinality of $X$ is at most $\mathfrak c$.

$X$ has a regular $G_\delta$-diagonal iff there is a collection of open sets of $X^2$, say $\lbrace U_n: n\in N\rbrace$, such that $\Delta=\bigcap\lbrace \overline{U_n}: n \in N\rbrace$, where $\Delta=\lbrace(x,x): x \in X\rbrace$.

Note that the question is answered; however I hope to get new proof.

By certain effort, If $|X|>\mathfrak c$, we can get an uncountable closed discrete subset $S$ of $X$, and for any point $x \in X$, there exists an open set $U_x$ such that $\overline{U_x} \cap S$ has at most one point.

I would like to know whether the following conjecture is right, wrong, or neither:

Let $X$ be a Hausdorff space. If $S \subset X$ is an uncountable closed discrete subset of $X$, and for any point $x \in X$, there exists an open set $U_x$ such that $\overline{U_x} \cap S$ has at most one point. Then could we obtain an uncountale collection of disjoint open sets in $X$?

Thanks for your any help.

I am struggling with this old problem, which is also posted here:

Let $X$ satisfy countable chain condition(abbreviated as CCC) and $X$ has a regular $G_\delta$-diagonal. Then the cardinality of $X$ is at most $\mathfrak c$.

$X$ has a regular $G_\delta$-diagonal iff there is a collection of open sets of $X^2$, say $\lbrace U_n: n\in N\rbrace$, such that $\Delta=\bigcap\lbrace \overline{U_n}: n \in N\rbrace$, where $\Delta=\lbrace(x,x): x \in X\rbrace$.

Note that the question is answered; however I hope to get new proof.

By certain effort, If $|X|>\mathfrak c$, we can get an uncountable closed discrete subset $S$ of $X$, and for any point $x \in X$, there exists an open set $U_x$ such that $\overline{U_x} \cap S$ has at most one point.

I would like to know whether the following conjecture is right, wrong, or neither:

Let $X$ be a Hausdorff space. If $S \subset X$ is an uncountable closed discrete subset of $X$, and for any point $x \in X$, there exists an open set $U_x$ such that $\overline{U_x} \cap S$ has at most one point. Then could we obtain an uncountale collection of disjoint open sets in $X$?

Thanks for your any help.

I am struggling with this old problem, which is also posted here:

Let $X$ satisfy countable chain condition(abbreviated as CCC) and $X$ has a regular $G_\delta$-diagonal. Then the cardinality of $X$ is at most $\mathfrak c$.

$X$ has a regular $G_\delta$-diagonal iff there is a collection of open sets of $X^2$, say $\lbrace U_n: n\in N\rbrace$, such that $\Delta=\bigcap\lbrace \overline{U_n}: n \in N\rbrace$, where $\Delta=\lbrace(x,x): x \in X\rbrace$.

Note that the question is answered; however I hope to get new proof.

By certain effort, If $|X|>\mathfrak c$, we can get an uncountable closed discrete subset $S$ of $X$, and for any point $x \in X$, there exists an open set $U_x$ such that $\overline{U_x} \cap S$ has at most one point.

I would like to know whether the following conjecture is right, wrong, or neither:

Let $X$ be a Hausdorff space. If $S \subset X$ is an uncountable closed discrete subset of $X$, and for any point $x \in X$, there exists an open set $U_x$ such that $\overline{U_x} \cap S$ has at most one point. Then could we obtain an uncountale collection of disjoint open sets in $X$?

Thanks for your any help.

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Paul
Paul
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A conjecture on closed discrete subset

I am struggling with this old problem, which is also posted here:

Let $X$ satisfy countable chain condition(abbreviated as CCC) and $X$ has a regular $G_\delta$-diagonal. Then the cardinality of $X$ is at most $\mathfrak c$.

$X$ has a regular $G_\delta$-diagonal iff there is a collection of open sets of $X^2$, say $\lbrace U_n: n\in N\rbrace$, such that $\Delta=\bigcap\lbrace \overline{U_n}: n \in N\rbrace$, where $\Delta=\lbrace(x,x): x \in X\rbrace$.

Note that the question is answered; however I hope to get new proof.

By certain effort, If $|X|>\mathfrak c$, we can get an uncountable closed discrete subset $S$ of $X$, and for any point $x \in X$, there exists an open set $U_x$ such that $\overline{U_x} \cap S$ has at most one point.

I would like to know whether the following conjecture is right, wrong, or neither:

Let $X$ be a Hausdorff space. If $S \subset X$ is an uncountable closed discrete subset of $X$, and for any point $x \in X$, there exists an open set $U_x$ such that $\overline{U_x} \cap S$ has at most one point. Then could we obtain an uncountale collection of disjoint open sets in $X$?

Thanks for your any help.