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added 125 characters in body

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edited Dec 5, 2009 at 10:01

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I'll risk making this a post, not a commment.

I think the real numbers $\mathbb R$ are a hemicompact $k$-space. Certainly $\mathbb R = \bigcup_n [-n,n]$ and if $K\subseteq\mathbb R$ is compact, then it's bounded, hence in some $[-n,n]$. It's a k-space, for if $K\subseteq\mathbb R$ has closed intersection with all compacts, then by looking at sequences, it's easy to see that $K$ is closed.

But $\mathbb R$ is not compact, so I guess you really mean to look at $C^b(\mathbb R)$, the algebra/space of all bounded continuous functions. Is that right? If not, then it's a whole new ball game (as $C(\mathbb R)$ the space of all continuous functions is not a Banach space).

But if so, then $C^b(\mathbb R)$ has character space $\beta\mathbb R$, and we can apply Jonas's construction: just pick a point $w\in\beta\mathbb R\setminus \mathbb R$ and evaluate there. This gives an algebra homomorphism $C^b(\mathbb R)\rightarrow\mathbb C$ which is not a composition operator.

Edit: Yes, the original question was about all continuous functions on X, not just the bounded ones. My mistake...

I'll risk making this a post, not a commment.

I think the real numbers $\mathbb R$ are a hemicompact $k$-space. Certainly $\mathbb R = \bigcup_n [-n,n]$ and if $K\subseteq\mathbb R$ is compact, then it's bounded, hence in some $[-n,n]$. It's a k-space, for if $K\subseteq\mathbb R$ has closed intersection with all compacts, then by looking at sequences, it's easy to see that $K$ is closed.

But $\mathbb R$ is not compact, so I guess you really mean to look at $C^b(\mathbb R)$, the algebra/space of all bounded continuous functions. Is that right? If not, then it's a whole new ball game (as $C(\mathbb R)$ the space of all continuous functions is not a Banach space).

But if so, then $C^b(\mathbb R)$ has character space $\beta\mathbb R$, and we can apply Jonas's construction: just pick a point $w\in\beta\mathbb R\setminus \mathbb R$ and evaluate there. This gives an algebra homomorphism $C^b(\mathbb R)\rightarrow\mathbb C$ which is not a composition operator.

I'll risk making this a post, not a commment.

I think the real numbers $\mathbb R$ are a hemicompact $k$-space. Certainly $\mathbb R = \bigcup_n [-n,n]$ and if $K\subseteq\mathbb R$ is compact, then it's bounded, hence in some $[-n,n]$. It's a k-space, for if $K\subseteq\mathbb R$ has closed intersection with all compacts, then by looking at sequences, it's easy to see that $K$ is closed.

But $\mathbb R$ is not compact, so I guess you really mean to look at $C^b(\mathbb R)$, the algebra/space of all bounded continuous functions. Is that right? If not, then it's a whole new ball game (as $C(\mathbb R)$ the space of all continuous functions is not a Banach space).

But if so, then $C^b(\mathbb R)$ has character space $\beta\mathbb R$, and we can apply Jonas's construction: just pick a point $w\in\beta\mathbb R\setminus \mathbb R$ and evaluate there. This gives an algebra homomorphism $C^b(\mathbb R)\rightarrow\mathbb C$ which is not a composition operator.

Edit: Yes, the original question was about all continuous functions on X, not just the bounded ones. My mistake...

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created Dec 4, 2009 at 9:00

18.7k
7
45
76

I'll risk making this a post, not a commment.

I think the real numbers $\mathbb R$ are a hemicompact $k$-space. Certainly $\mathbb R = \bigcup_n [-n,n]$ and if $K\subseteq\mathbb R$ is compact, then it's bounded, hence in some $[-n,n]$. It's a k-space, for if $K\subseteq\mathbb R$ has closed intersection with all compacts, then by looking at sequences, it's easy to see that $K$ is closed.

But $\mathbb R$ is not compact, so I guess you really mean to look at $C^b(\mathbb R)$, the algebra/space of all bounded continuous functions. Is that right? If not, then it's a whole new ball game (as $C(\mathbb R)$ the space of all continuous functions is not a Banach space).

But if so, then $C^b(\mathbb R)$ has character space $\beta\mathbb R$, and we can apply Jonas's construction: just pick a point $w\in\beta\mathbb R\setminus \mathbb R$ and evaluate there. This gives an algebra homomorphism $C^b(\mathbb R)\rightarrow\mathbb C$ which is not a composition operator.