MHB Unbounded subset of ordinals a set?

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An unbounded subset C of the class of all ordinals R cannot be a set, but rather a class. This conclusion arises from the ability to establish a one-to-one correspondence between C and R, which implies that if C were a set, it would be bijective with some ordinal A. However, this leads to a contradiction since A cannot be bijective with R. The discussion assumes the framework of Zermelo-Fraenkel set theory with the Axiom of Choice (ZFC). Therefore, the assertion that an unbounded subset of ordinals is a class rather than a set is supported by this reasoning.
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Let R be the class of all ordinals. If a subset C of R is unbounded (i.e. for any ordinal \alpha \in R, there is \beta in C with \beta greater than \alpha ), then it seems to me that C cannot be a set, only a class. Is this true, and if so, how does one prove it? My reading on the general subject matter is limited to a bit of web browsing - perhaps the problem is trivial.
 
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RWood said:
Let R be the class of all ordinals. If a subset C of R is unbounded (i.e. for any ordinal \alpha \in R, there is \beta in C with \beta greater than \alpha ), then it seems to me that C cannot be a set, only a class. Is this true, and if so, how does one prove it? My reading on the general subject matter is limited to a bit of web browsing - perhaps the problem is trivial.

I think I have the outline of a proof (there may of course be something much quicker!).

1) It is quite easy to get a 1-1 correspondence between C and R; a map C=>R is obvious; a 1-1 map R=>C can be constructed by transfinite induction, using
the unboundedness of C to ensure successor elements (or limit ordinals) are mapped to an increasing sequence of C-members.

2) On the other hand, if C is a set then it is bijective with some ordinal A (and some cardinal as well). But then A would be bijective with R, and that is clearly impossible. All this assumes we are a ZFC world.
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...

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