In $C_p$-Theory monographs, it is said that two topological spaces $X$ and $Y$ are said to be $t$-equivalent means that $C_p(X)$ is homeomorphic to $C_p(Y)$. Then they also define $u$-equivalences (for uniform spaces $C_p(X)$ and $C_p(Y)$), i want to know can we define the same way $t$-equivalence for any other topology on $C(X)$ and $C(Y)$ ?
$\begingroup$
$\endgroup$
6
-
$\begingroup$ It's not clear to me: What do you mean by uniform space $C_p(X)$? Of course these can be considered as tvs, hence as uniform space. I think you mean that the function spaces are uniformly equivalent (not necessarily homeomorphic). And further, you can of course consider other function spaces as well, f.i. $C_b(X)$, ...You always get an equivalence relation. $\endgroup$– Dieter KadelkaCommented Apr 10, 2022 at 13:40
-
$\begingroup$ If you have a uniform way of assigning a topology to $C(X)$ then you can define an accompanying notion of equivalence. It would be prudent to use different names for these; $t$-equivalence is connected to the topology of pointwise convergence. You could say, for example, that $X$ and $Y$ are $co$-equivalent if $C(X)$ and $C(Y)$ are homeomorphic when endowed with the compact-open topology. $\endgroup$– KP HartCommented Apr 15, 2022 at 8:21
-
$\begingroup$ @KPHart Does that mean we can define the equivalence for $C(X)$ and $C(Y)$ endowed with any topology? Let's talk about $m$-topology, can we define the m-equivalence(say) the same way we define for topology of pointwise convergence? $\endgroup$– Mir AaliyaCommented Apr 16, 2022 at 6:50
-
$\begingroup$ @MirAaliya Yes, every such uniform way yields a notion of equivalence. $\endgroup$– KP HartCommented Apr 16, 2022 at 11:40
-
$\begingroup$ @KPHart Thank you so much sir. One more thing I want to know is that what is the immediate application of defining such equivalences for any topology on $C(X)$ and $C(Y)$. What I feel is t-equivalence works for topology on $C(X)$ and $C(Y)$, like we also have u-equivalence which works for uniformity on the same. Am I mistaken or I got it right ? $\endgroup$– Mir AaliyaCommented Apr 17, 2022 at 6:29
|
Show 1 more comment