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Piotr Hajlasz
Piotr Hajlasz
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The answer is in the negative and a construction of a counterexample is provided here. I wrote this argument quickly, but I hope it is correct. Let me know if it is not. We will construct a family of lines passing through the origin in $\mathbb{R}^3$ so that:

1. Every plane passing through the origin contains at least one of the lines from the family,

2. The union of the lines does not contain any plane passing through the origin.

The shortest way to do it is to use transfinite induction. While this is not an effective construction, it shows that you cannot prove your claim and this is what is important here. The argument used here is pretty similar to the one used in this post.

Order all planes by the initial ordinal of $2^{\aleph_0}$. Denote the planes by $P_\alpha$, where $\alpha$ is an ordinal. We can construct a corresponding family of lines $\ell_\alpha$ so that

(a) $\ell_\alpha\subset P_\alpha$,

(b) $\ell_\alpha\cap P_\beta=\{ 0\}$ for $\beta<\alpha$.

Then (1) is obviously satisfied by (a). By (b), each plane $P_\beta$ can contain only lines $\ell_\alpha$, $\alpha\leq\beta$. The cardinality of such lines is less than $2^{\aleph_0}$ which is the cardinality of all lines in a plane so (2) follows.

The construction of the family of lines $\ell_\alpha$ is pretty standard. Suppose we already have lines $\ell_\beta$, $\beta<\alpha$ satisfying (1) and (2). Consider the plane $P_\alpha$. $P_\alpha\cap P_\beta$, $\beta<\alpha$ is a line and the cardinality of such lines is less than $2^{\aleph_0}$ so there is a line $\ell_\alpha\subset P_\alpha$ which is different from all lines $P_\alpha\cap P_\beta$. Hence it is not contained in any of the planes $P_\beta$, $\beta<\alpha$ which is condition (2).

I wrote this argument quickly, but I hope it is correct. Let me know if it is not. We will construct a family of lines passing through the origin in $\mathbb{R}^3$ so that:

1. Every plane passing through the origin contains at least one of the lines from the family,

2. The union of the lines does not contain any plane passing through the origin.

The shortest way to do it is to use transfinite induction. While this is not an effective construction, it shows that you cannot prove your claim and this is what is important here. The argument used here is pretty similar to the one used in this post.

Order all planes by the initial ordinal of $2^{\aleph_0}$. Denote the planes by $P_\alpha$, where $\alpha$ is an ordinal. We can construct a corresponding family of lines $\ell_\alpha$ so that

(a) $\ell_\alpha\subset P_\alpha$,

(b) $\ell_\alpha\cap P_\beta=\{ 0\}$ for $\beta<\alpha$.

Then (1) is obviously satisfied by (a). By (b), each plane $P_\beta$ can contain only lines $\ell_\alpha$, $\alpha\leq\beta$. The cardinality of such lines is less than $2^{\aleph_0}$ which is the cardinality of all lines in a plane so (2) follows.

The construction of the family of lines $\ell_\alpha$ is pretty standard. Suppose we already have lines $\ell_\beta$, $\beta<\alpha$ satisfying (1) and (2). Consider the plane $P_\alpha$. $P_\alpha\cap P_\beta$, $\beta<\alpha$ is a line and the cardinality of such lines is less than $2^{\aleph_0}$ so there is a line $\ell_\alpha\subset P_\alpha$ which is different from all lines $P_\alpha\cap P_\beta$. Hence it is not contained in any of the planes $P_\beta$, $\beta<\alpha$ which is condition (2).

The answer is in the negative and a construction of a counterexample is provided here. I wrote this argument quickly, but I hope it is correct. Let me know if it is not. We will construct a family of lines passing through the origin in $\mathbb{R}^3$ so that:

1. Every plane passing through the origin contains at least one of the lines from the family,

2. The union of the lines does not contain any plane passing through the origin.

The shortest way to do it is to use transfinite induction. While this is not an effective construction, it shows that you cannot prove your claim and this is what is important here. The argument used here is pretty similar to the one used in this post.

Order all planes by the initial ordinal of $2^{\aleph_0}$. Denote the planes by $P_\alpha$, where $\alpha$ is an ordinal. We can construct a corresponding family of lines $\ell_\alpha$ so that

(a) $\ell_\alpha\subset P_\alpha$,

(b) $\ell_\alpha\cap P_\beta=\{ 0\}$ for $\beta<\alpha$.

Then (1) is obviously satisfied by (a). By (b), each plane $P_\beta$ can contain only lines $\ell_\alpha$, $\alpha\leq\beta$. The cardinality of such lines is less than $2^{\aleph_0}$ which is the cardinality of all lines in a plane so (2) follows.

The construction of the family of lines $\ell_\alpha$ is pretty standard. Suppose we already have lines $\ell_\beta$, $\beta<\alpha$ satisfying (1) and (2). Consider the plane $P_\alpha$. $P_\alpha\cap P_\beta$, $\beta<\alpha$ is a line and the cardinality of such lines is less than $2^{\aleph_0}$ so there is a line $\ell_\alpha\subset P_\alpha$ which is different from all lines $P_\alpha\cap P_\beta$. Hence it is not contained in any of the planes $P_\beta$, $\beta<\alpha$ which is condition (2).

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Piotr Hajlasz
Piotr Hajlasz
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I think therewrote this argument quickly, but I hope it is a counterexamplecorrect. I am typing details of a construction ofLet me know if it is not. We will construct a setfamily of lines in $\mathbb{R}^3$ passing through the origin suchin $\mathbb{R}^3$ so that every plane passing through:

1. Every plane passing through the origin contains at least one of the lines from the family,

2. The union of the lines does not contain any plane passing through the origin.

The shortest way to do it is to use transfinite induction. While this is not an effective construction, it shows that you cannot prove your claim and this is what is important here. The argument used here is pretty similar to the origin contains positive but finite numberone used in this post.

Order all planes by the initial ordinal of $2^{\aleph_0}$. Denote the planes by $P_\alpha$, where $\alpha$ is an ordinal. We can construct a corresponding family of lines $\ell_\alpha$ so that

(a) $\ell_\alpha\subset P_\alpha$,

(b) $\ell_\alpha\cap P_\beta=\{ 0\}$ for $\beta<\alpha$.

Then (1) is obviously satisfied by (a). By (b), each plane $P_\beta$ can contain only lines $\ell_\alpha$, $\alpha\leq\beta$. The cardinality of

such lines is less than $2^{\aleph_0}$ which is the cardinality of all lines in a plane so (2) follows. I hope my

The

construction of the family of lines $\ell_\alpha$ is correctpretty standard. Don't vote on my answer until I provide detailsSuppose we already have lines $\ell_\beta$, $\beta<\alpha$ satisfying (1) and (2). Wait forConsider the updateplane $P_\alpha$. $P_\alpha\cap P_\beta$, $\beta<\alpha$ is a line and the cardinality of such lines is less than $2^{\aleph_0}$ so there is a line $\ell_\alpha\subset P_\alpha$ which is different from all lines $P_\alpha\cap P_\beta$. Hence it is not contained in any of the planes $P_\beta$, $\beta<\alpha$ which is condition (2).

  

I think there is a counterexample. I am typing details of a construction of a set of lines in $\mathbb{R}^3$ passing through the origin such that every plane passing through the origin contains positive but finite number of such lines. I hope my construction is correct. Don't vote on my answer until I provide details. Wait for the update.

 

I wrote this argument quickly, but I hope it is correct. Let me know if it is not. We will construct a family of lines passing through the origin in $\mathbb{R}^3$ so that:

1. Every plane passing through the origin contains at least one of the lines from the family,

2. The union of the lines does not contain any plane passing through the origin.

The shortest way to do it is to use transfinite induction. While this is not an effective construction, it shows that you cannot prove your claim and this is what is important here. The argument used here is pretty similar to the one used in this post.

Order all planes by the initial ordinal of $2^{\aleph_0}$. Denote the planes by $P_\alpha$, where $\alpha$ is an ordinal. We can construct a corresponding family of lines $\ell_\alpha$ so that

(a) $\ell_\alpha\subset P_\alpha$,

(b) $\ell_\alpha\cap P_\beta=\{ 0\}$ for $\beta<\alpha$.

Then (1) is obviously satisfied by (a). By (b), each plane $P_\beta$ can contain only lines $\ell_\alpha$, $\alpha\leq\beta$. The cardinality of

such lines is less than $2^{\aleph_0}$ which is the cardinality of all lines in a plane so (2) follows.

The

construction of the family of lines $\ell_\alpha$ is pretty standard. Suppose we already have lines $\ell_\beta$, $\beta<\alpha$ satisfying (1) and (2). Consider the plane $P_\alpha$. $P_\alpha\cap P_\beta$, $\beta<\alpha$ is a line and the cardinality of such lines is less than $2^{\aleph_0}$ so there is a line $\ell_\alpha\subset P_\alpha$ which is different from all lines $P_\alpha\cap P_\beta$. Hence it is not contained in any of the planes $P_\beta$, $\beta<\alpha$ which is condition (2).

 
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Piotr Hajlasz
Piotr Hajlasz
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I think there is a counterexample. I am typing details of a construction of a set of lines in $\mathbb{R}^3$ passing through the origin such that every plane passing through the origin contains positive but finite number of such lines. I hope my construction is correct. Don't vote on my answer until I provide details. Wait for the update.