Let $X$ be a Tychonof space and $\beta X$ is its compactification. Then could $I^X$ be seen as a subspace of $I^{\beta X}$ under the compact-open topology?
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$\begingroup$ What do you mean bu $I^X$? The set of all continuous functions (as in Engelking's book) or the set of all functions? $\endgroup$– KP HartCommented Jan 9, 2012 at 11:48
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1$\begingroup$ Only continuous functions in the $I^X$:] $\endgroup$– PaulCommented Jan 9, 2012 at 13:13
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1 Answer
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The two sets are essentially the same: the map that sends every $f\in I^{\beta X}$ to its restriction is a bijection; the two topologies are, in general, not the same. The compact-open topology on $I^{\mathbb{N}}$ is the product topology, whereas the compact-open topology on $I^{\beta\mathbb{N}}$ is the topology induced by the uniform metric. In fact: on $I^{\beta X}$ the compact-open topology is the uniform topology (this only needs compactness of the domain space). On the other hand, the compact-open topology on $I^X$ is strictly weaker than the uniform topology (if $X$ is not compact): the set $U=\lbrace f:\sup_x|f(x)|<\frac12\rbrace$ is open in the uniform topology but not in the compact-open topology: if $O$ is a basic open set determined by compact sets $K_1,\ldots,K_n$ and open sets $U_1,\ldots,U_n$ then one can take a point $x$ outside $\bigcup_{i=1}^nK_i$ and construct an $f$ in $O$ that satisfies $f(x)=1$.
