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Dejan Govc
Dejan Govc
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I believe such a space cannot exist for the following reason:

Suppose it does. By the second requirement, there should be an unbounded function $f: X \to (0, \infty)$, which is bounded on every compact set. This means we have countably many points $x_1, x_2, x_3, \dots$ such that for each $n$ the inequality $f(x_n)\ge n$ holds. The singletons $\lbrace x_n\rbrace$ are finite sets, therefore compact. By the first requirement, the set $\lbrace x_n| n\in\mathbb{N}\rbrace$ is therefore relatively compact and so its closure must be compact. But then f$f$ is unbounded on a compact set, which is a contradiction.

I believe such a space cannot exist for the following reason:

Suppose it does. By the second requirement, there should be an unbounded function $f: X \to (0, \infty)$, which is bounded on every compact set. This means we have countably many points $x_1, x_2, x_3, \dots$ such that for each $n$ the inequality $f(x_n)\ge n$ holds. The singletons $\lbrace x_n\rbrace$ are finite sets, therefore compact. By the first requirement, the set $\lbrace x_n| n\in\mathbb{N}\rbrace$ is therefore relatively compact and so its closure must be compact. But then f is unbounded on a compact set, which is a contradiction.

I believe such a space cannot exist for the following reason:

Suppose it does. By the second requirement, there should be an unbounded function $f: X \to (0, \infty)$, which is bounded on every compact set. This means we have countably many points $x_1, x_2, x_3, \dots$ such that for each $n$ the inequality $f(x_n)\ge n$ holds. The singletons $\lbrace x_n\rbrace$ are finite sets, therefore compact. By the first requirement, the set $\lbrace x_n| n\in\mathbb{N}\rbrace$ is therefore relatively compact and so its closure must be compact. But then $f$ is unbounded on a compact set, which is a contradiction.

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Dejan Govc
Dejan Govc
  • 558
  • 1
  • 4
  • 15

I believe such a space cannot exist for the following reason:

Suppose it does. By the second requirement, there should be an unbounded function $f: X \to (0, \infty)$, which is bounded on every compact set. This means we have countably many points $x_1, x_2, x_3, \dots$ such that for each $n$ the inequality $f(x_n)\ge n$ holds. The singletons {$x_n$}$\lbrace x_n\rbrace$ are finite sets, therefore compact. By the first requirement, the set {$x_n| n\in\mathbb{N}$}$\lbrace x_n| n\in\mathbb{N}\rbrace$ is therefore relatively compact and so its closure must be compact. But then f is unbounded on a compact set, which is a contradiction.

I believe such a space cannot exist for the following reason:

Suppose it does. By the second requirement, there should be an unbounded function $f: X \to (0, \infty)$, which is bounded on every compact set. This means we have countably many points $x_1, x_2, x_3, \dots$ such that for each $n$ the inequality $f(x_n)\ge n$ holds. The singletons {$x_n$} are finite sets, therefore compact. By the first requirement, the set {$x_n| n\in\mathbb{N}$} is therefore relatively compact and so its closure must be compact. But then f is unbounded on a compact set, which is a contradiction.

I believe such a space cannot exist for the following reason:

Suppose it does. By the second requirement, there should be an unbounded function $f: X \to (0, \infty)$, which is bounded on every compact set. This means we have countably many points $x_1, x_2, x_3, \dots$ such that for each $n$ the inequality $f(x_n)\ge n$ holds. The singletons $\lbrace x_n\rbrace$ are finite sets, therefore compact. By the first requirement, the set $\lbrace x_n| n\in\mathbb{N}\rbrace$ is therefore relatively compact and so its closure must be compact. But then f is unbounded on a compact set, which is a contradiction.

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Dejan Govc
Dejan Govc
  • 558
  • 1
  • 4
  • 15

I believe such a space cannot exist for the following reason:

Suppose it does. By the second requirement, there should be an unbounded function $f: X \to (0, \infty)$, which is bounded on every compact set. This means we have countably many points $x_1, x_2, x_3, \dots$ such that for each $n$ the inequality $f(x_n)\ge n$ holds. The singletons {$x_n$} are finite sets, therefore compact. By the first requirement, the set {$x_n| n\in\mathbb{N}$} is therefore relatively compact and so its closure must be compact. But then f is unbounded on a compact set, which is a contradiction.