Could $I^X$ be seen as a subspace of $I^{\beta X}$ under the compact-open topology? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T20:58:40Z http://mathoverflow.net/feeds/question/85017 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85017/could-ix-be-seen-as-a-subspace-of-i-beta-x-under-the-compact-open-topolog Could $I^X$ be seen as a subspace of $I^{\beta X}$ under the compact-open topology? Paul 2012-01-06T00:47:25Z 2012-01-10T07:30:30Z <p>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?</p> http://mathoverflow.net/questions/85017/could-ix-be-seen-as-a-subspace-of-i-beta-x-under-the-compact-open-topolog/85292#85292 Answer by KP Hart for Could $I^X$ be seen as a subspace of $I^{\beta X}$ under the compact-open topology? KP Hart 2012-01-09T22:17:25Z 2012-01-10T07:30:30Z <p>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 $U=\lbrace f:\sup_x|f(x)|&lt;\frac12\rbrace$ 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 $f(x)=1$. </p>