# Isomorphic Structures

• November 7th 2012, 01:38 PM
drg
Isomorphic Structures
Hi everyone,

This problem is about first order logic.

Given a structure A and a bijection g with domain |A| (i.e. the universe of A), I have to show that there is a unique structure B such that g is an isomorphism of A onto B.

Since A is isomorphic to itself, this basically comes down to showing that every structure is unique up to isomorphism. While I would feel somewhat comfortable showing this for certain types of sets, I'm having trouble seeing how to go about doing this with structure.

Any help would be appreciated.
• November 7th 2012, 02:08 PM
Plato
Re: Isomorphic Structures
Quote:

Originally Posted by drg
HGiven a structure A and a bijection g with domain |A|, I have to show that there is a unique structure B such that g is an isomorphism of A onto B.
.

..a structure A what does that term mean?
• November 8th 2012, 01:23 AM
emakarov
Re: Isomorphic Structures
Quote:

Originally Posted by drg
Given a structure A and a bijection g with domain |A| (i.e. the universe of A), I have to show that there is a unique structure B such that g is an isomorphism of A onto B.

I believe this is a better link that explains what a structure is in the context of mathematical logic.

If I understand correctly, g is an isomorphism between A and B if

(1) g is a bijection between |A| and |B|,
(2) for every functional symbol f and every $\vec{a}\in |A|$ it is the case that $g(f^A(\vec{a}))=f^B(g(\vec{a}))$, and
(3) for every predicate symbol R and every $\vec{a}\in |A|$ it is the case that $R^A(\vec{a})$ iff $R^B(g(\vec{a}))$.

The universe of B must be the image of g. We need to show that there exists one and only one interpretation of functional and predicate symbols on |B| such that g is an isomorphism. Well, this interpretation is uniquely determined by properties (2) and (3).
• November 8th 2012, 05:49 PM
drg
Re: Isomorphic Structures
Quote:

Originally Posted by emakarov
I believe this is a better link that explains what a structure is in the context of mathematical logic.

If I understand correctly, g is an isomorphism between A and B if

(1) g is a bijection between |A| and |B|,
(2) for every functional symbol f and every $\vec{a}\in |A|$ it is the case that $g(f^A(\vec{a}))=f^B(g(\vec{a}))$, and
(3) for every predicate symbol R and every $\vec{a}\in |A|$ it is the case that $R^A(\vec{a})$ iff $R^B(g(\vec{a}))$.

The universe of B must be the image of g. We need to show that there exists one and only one interpretation of functional and predicate symbols on |B| such that g is an isomorphism. Well, this interpretation is uniquely determined by properties (2) and (3).