(08/30/2012, 05:24 PM)JmsNxn Wrote: Well. The reason I ask is because I was structuring my semi operators around the distribution of the set:Sorry if my question is stupid (I'm a newbie in math) but I'm a bit confused.
\( \mathbb{I}_{y} = \{ s_0 | s_0 \in \mathbb{R}\,\,;\,\,x\,\,\bigtriangleup_{s_0}\,\,y\,\,\in \mathbb{N}} \)
claim that there are operators unique to x and y which allow us to perform operations on elements of \( \mathbb{I}_y \) instead of operations on \( \mathbb{N} \). We then say that \( n\,\,\bigtriangleup_{s_n}\,\,y \) is an isomorphism from \( \mathbb{I}_y \to \mathbb{N} \)
From your definition of \( \mathbb{I}_{y} \) (that is a set of the real ranks \( s_0 \) that satysfie \( \,\,x\,\,\bigtriangleup_{s_0}\,\,y\,\,\in \mathbb{N} \) for a fixed \( y \) ) so we have that \( \mathbb{I}_{y}\subset \mathbb R \).
In other words we can define a function \( I_y:\mathbb {N} \rightarrow R \) whit the property \( I_y(n)=s_n \) such that \( n\,\,\bigtriangleup_{s_n}\,\,y\in \mathbb{N} \), at this point we have that the set of the \( s_0, s_1, s_3, ... \) is your \( \mathbb{I}_{y} \)
The question is, how do you know that \( I_y:\mathbb {N} \rightarrow R \) is injective? It can be maybe a surjection on the reals (not-injective, multivalued)?
In fact from your definiton of \( \mathbb{I}_{y} \) seems me that you except it to be a countable subset of \( \mathbb{R} \) then you are asuming (is an hypothesis?) that \( I_y:\mathbb {N} \rightarrow \mathbb{I}_{y} \) is a bijection (is this what you mean when you say that they are isomorphic? And which is the funtion you use for the isomorphism? maybe you use \( I^{\circ -1}_y:\mathbb{I}_{y} \rightarrow \mathbb {N} \) ? )
Thanks in advance, and sorry for my bad english.
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)
