Return to Answer

Any metric space can be isometrically embedded into some Banach space E. And E has the property X, since it can be topologically embedded into itself as its open unit ball, which is clearly not closed in E. Therefore, the answer to 3. is YES, and the intuition of Robin Saunders is very good.

This is just to clarify some things. For this, let $(M,d)$ be a complete metric space.

It is pretty clear that a compact metric space cannot have "the property X". Conversely, asumming that $M$ is non-compact, by taking into consideration the equivalent metric $d^*(x,y)$ := $\min$ { $d(x,y)$, $f(x)+f(y)$ }, where $f:M$ $\rightarrow$ $(0,\infty)$ is continuous and inf$_{M}$ $f$ = 0, one easily obtain that $(M,d^*)$ is not complete.

Consequently, $M$ has "the property X" iff it is non-compact.

Now, every separable metric space can be embedded into the Hilbert cube, which has not "the property X" .

[Note that on Wikipedia you'll find "every separable metric space can be isometrically embedded into the Hilbert cube", which is obviously not true.]

OTOH, any subspace of a separable metric space is separable, too. And any compact metric space is, clearly, separable.

Therefore, $M$ is embeddable into a space having not "the property X" iff $M$ is separable.

Any metric space can be isometrically embedded into some Banach space E. And E has the property X, since it can be topologically embedded into itself as its open unit ball, which is clearly not closed in E. Therefore, the answer to 3. is YES, and the intuition of Robin Saunders is very good.

This is just to clarify some things. For this, let $(M,d)$ be a complete metric space.

It is pretty clear that a compact metric space cannot have "the property X". Conversely, asumming that $M$ is non-compact and that $d$ is bounded, by taking into consideration the equivalent metric $d^{*}\left(x,y\right)$= $\sup_{t\in M}$ $\mid f(x)\cdot d(x,t)$ $-f\left(y\right)\cdot d\left(y,t\right)\mid$, where $f:M$ $\rightarrow$ $(0,\infty)$ is bounded and continuous and inf$_{M}$ $f$ = 0, one easily obtain that $(M,d^*)$ is not complete.

Consequently, $M$ has "the property X" iff it is non-compact.

Now, every separable metric space can be embedded into the Hilbert cube, which has not "the property X" .

[Note that on Wikipedia you'll find "every separable metric space can be isometrically embedded into the Hilbert cube", which is obviously not true.]

OTOH, any subspace of a separable metric space is separable, too. And any compact metric space is, clearly, separable.

Therefore, $M$ is embeddable into a space having not "the property X" iff $M$ is separable.

Any metric space can be isometrically embedded into some Banach space E. And E has the property X, since it can be topologically embedded into itself as its open unit ball, which is clearly not closed in E. Therefore, the answer to 3. is YES, and the intuition of Robin Saunders is very good.

This is just to clarify some things. For this, let $(M,d)$ be a complete metric space.

It is pretty clear that a compact metric space cannot have "the property X". Conversely, asumming that $M$ is non-compact, by taking into consideration the equivalent metric $d^*(x,y)$ := $\min$ { $d(x,y)$, $f(x)+f(y)$ }, where $f:M$ $\rightarrow$ $(0,\infty)$ is continuous and inf$_{M}$ $f$ = 0, one easily obtain that $(M,d^*)$ is not complete.

Consequently, $M$ has "the property X" iff it is compact.

Now, every separable metric space can be embedded into the Hilbert cube, which has not "the property X" .

[Note that on Wikipedia you'll find "every separable metric space can be isometrically embedded into the Hilbert cube", which is obviously not true.]

OTOH, any subspace of a separable metric space is separable, too. And any compact metric space is, clearly, separable.

Therefore, $M$ is embeddable into a space having not "the property X" iff $M$ is separable.

Any metric space can be isometrically embedded into some Banach space E. And E has the property X, since it can be topologically embedded into itself as its open unit ball, which is clearly not closed in E. Therefore, the answer to 3. is YES, and the intuition of Robin Saunders is very good.

This is just to clarify some things. For this, let $(M,d)$ be a complete metric space.

It is pretty clear that a compact metric space cannot have "the property X". Conversely, asumming that $M$ is non-compact, by taking into consideration the equivalent metric $d^*(x,y)$ := $\min$ { $d(x,y)$, $f(x)+f(y)$ }, where $f:M$ $\rightarrow$ $(0,\infty)$ is continuous and inf$_{M}$ $f$ = 0, one easily obtain that $(M,d^*)$ is not complete.

Consequently, $M$ has "the property X" iff it is non-compact.

Now, every separable metric space can be embedded into the Hilbert cube, which has not "the property X" .

[Note that on Wikipedia you'll find "every separable metric space can be isometrically embedded into the Hilbert cube", which is obviously not true.]

OTOH, any subspace of a separable metric space is separable, too. And any compact metric space is, clearly, separable.

Therefore, $M$ is embeddable into a space having not "the property X" iff $M$ is separable.

Any metric space can be isometrically embedded into some Banach space E. And E has the property X, since it can be topologically embedded into itself as its open unit ball, which is clearly not closed in E. Therefore, the answer to 3. is YES, and the intuition of Robin Saunders is very good.

Any metric space can be isometrically embedded into some Banach space E. And E has the property X, since it can be topologically embedded into itself as its open unit ball, which is clearly not closed in E. Therefore, the answer to 3. is YES, and the intuition of Robin Saunders is very good.

This is just to clarify some things. For this, let $(M,d)$ be a complete metric space.

It is pretty clear that a compact metric space cannot have "the property X". Conversely, asumming that $M$ is non-compact, by taking into consideration the equivalent metric $d^*(x,y)$ := $\min$ { $d(x,y)$, $f(x)+f(y)$ }, where $f:M$ $\rightarrow$ $(0,\infty)$ is continuous and inf$_{M}$ $f$ = 0, one easily obtain that $(M,d^*)$ is not complete.

Consequently, $M$ has "the property X" iff it is compact.

Now, every separable metric space can be embedded into the Hilbert cube, which has not "the property X" .

[Note that on Wikipedia you'll find "every separable metric space can be isometrically embedded into the Hilbert cube", which is obviously not true.]

OTOH, any subspace of a separable metric space is separable, too. And any compact metric space is, clearly, separable.

Therefore, $M$ is embeddable into a space having not "the property X" iff $M$ is separable.