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Not a complete answer, feel free to edit. (Likely, the answer is not known anyway.) The distance is at least 2. Look at both spaces as $C(K)$ spaces. Corresponding $K$'s are not homeomorphic, see the discussion here http://math.stackexchange.com/questions/207435/isometry-between-l-infty-and-ell-inftyhttps://math.stackexchange.com/questions/207435/isometry-between-l-infty-and-ell-infty

So by Amir-Cambern theorem (near isometry property of $C(K)$ spaces) the distance is at least 2. The theorem says if there is an isomorphism between $C(K_1)$ and $C(K_2)$ with distortion strictly less than 2, then $K_1$ and $K_2$ are homeomorphic.

Not a complete answer, feel free to edit. (Likely, the answer is not known anyway.) The distance is at least 2. Look at both spaces as $C(K)$ spaces. Corresponding $K$'s are not homeomorphic, see the discussion here http://math.stackexchange.com/questions/207435/isometry-between-l-infty-and-ell-infty

So by Amir-Cambern theorem (near isometry property of $C(K)$ spaces) the distance is at least 2. The theorem says if there is an isomorphism between $C(K_1)$ and $C(K_2)$ with distortion strictly less than 2, then $K_1$ and $K_2$ are homeomorphic.

Not a complete answer, feel free to edit. (Likely, the answer is not known anyway.) The distance is at least 2. Look at both spaces as $C(K)$ spaces. Corresponding $K$'s are not homeomorphic, see the discussion here https://math.stackexchange.com/questions/207435/isometry-between-l-infty-and-ell-infty

So by Amir-Cambern theorem (near isometry property of $C(K)$ spaces) the distance is at least 2. The theorem says if there is an isomorphism between $C(K_1)$ and $C(K_2)$ with distortion strictly less than 2, then $K_1$ and $K_2$ are homeomorphic.

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Bunyamin Sari
Bunyamin Sari
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Not a complete answer, feel free to edit. (Likely, the answer is not known anyway.) The distance is at least 2. Look at both spaces as $C(K)$ spaces. In the case of $\ell_{\infty}$, $K=\beta\mathbb N$. I do not know whatCorresponding $K$ is for $L_{\infty}$ (does anyone know?) but cannot be's are not homeomorphic to $\beta\mathbb N$ because otherwise the spaces would be isometric. We know that $\ell_{\infty}$ cannot be isometric to $L_{\infty}$. Because the cardinality of, see the extreme points of $L_{\infty}$ is larger than that of $\ell_{\infty}$. (I think there are more functions on $[0,1]$ taking values 1 a.e. than infinitediscussion here $\pm$1 sequences, correct?)http://math.stackexchange.com/questions/207435/isometry-between-l-infty-and-ell-infty

So by Amir-Cambern theorem (near isometry property of $C(K)$ spaces) the distance is at least 2. The theorem says if there is an isomorphism between $C(K_1)$ and $C(K_2)$ with distortion strictly less than 2, then $K_1$ and $K_2$ are homeomorphic.

Not a complete answer, feel free to edit. (Likely, the answer is not known anyway.) The distance is at least 2. Look at both spaces as $C(K)$ spaces. In the case of $\ell_{\infty}$, $K=\beta\mathbb N$. I do not know what $K$ is for $L_{\infty}$ (does anyone know?) but cannot be homeomorphic to $\beta\mathbb N$ because otherwise the spaces would be isometric. We know that $\ell_{\infty}$ cannot be isometric to $L_{\infty}$. Because the cardinality of the extreme points of $L_{\infty}$ is larger than that of $\ell_{\infty}$. (I think there are more functions on $[0,1]$ taking values 1 a.e. than infinite $\pm$1 sequences, correct?) So by Amir-Cambern theorem (near isometry property of $C(K)$ spaces) the distance is at least 2. The theorem says if there is an isomorphism between $C(K_1)$ and $C(K_2)$ with distortion strictly less than 2, then $K_1$ and $K_2$ are homeomorphic.

Not a complete answer, feel free to edit. (Likely, the answer is not known anyway.) The distance is at least 2. Look at both spaces as $C(K)$ spaces. Corresponding $K$'s are not homeomorphic, see the discussion here http://math.stackexchange.com/questions/207435/isometry-between-l-infty-and-ell-infty

So by Amir-Cambern theorem (near isometry property of $C(K)$ spaces) the distance is at least 2. The theorem says if there is an isomorphism between $C(K_1)$ and $C(K_2)$ with distortion strictly less than 2, then $K_1$ and $K_2$ are homeomorphic.

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Bunyamin Sari
Bunyamin Sari
  • 2.4k
  • 14
  • 14

Not a complete answer, feel free to edit. (Likely, the answer is not known anyway.) The distance is at least 2. Look at both spaces as $C(K)$ spaces. In the case of $\ell_{\infty}$, $K=\beta\mathbb N$. I do not know what $K$ is for $L_{\infty}$ (does anyone know?) but cannot be homeomorphic to $\beta\mathbb N$ because otherwise the spaces would be isometric. We know that $\ell_{\infty}$ cannot be isometric to $L_{\infty}$. Because the cardinality of the extreme points of $L_{\infty}$ is larger than that of $\ell_{\infty}$. (I think there are more functions on $[0,1]$ taking values 1 a.e. than infinite $\pm$1 sequences, correct?) So by Amir-Cambern theorem (near isometry property of $C(K)$ spaces) the distance is at least 2. The theorem says if there is an isomorphism between $C(K_1)$ and $C(K_2)$ with distortion strictly less than 2, then $K_1$ and $K_2$ are homeomorphic.