Is there always isomorphism between two sets that has the same cardinality ?
We only know that the two sets have the same cardinality. I tried to find a counter example but couldn't.
[Math] lways isomorphism between two sets that have the same cardinality
cardinalselementary-set-theory
Related Solutions
The definition of when sets $X$ and $Y$ have the same cardinality is that there exists a function $f:X\to Y$ which is both one-to-one and onto. So according to the definition, you need a single function with both properties at once.
However, it turns out that if you have separate functions $X\to Y$ with just one of the properties each, this implies there exists a function with both at once. Specifically, suppose $g:X\to Y$ is one-to-one and $h:X\to Y$ is onto. Take a function $i:Y\to X$ which is a sort of "inverse" to $h$, in the sense that for each $y\in Y$, $i(y)$ is one of the points $x\in X$ such that $h(x)=y$ (we know that at least one such $x$ exists since $h$ is onto). That is, we have $h(i(y))=y$ for all $y\in Y$.
I claim now that $i:Y\to X$ is one-to-one. Indeed, suppose that $i(y)=i(y')$. Applying $h$ to both sides, we get $h(i(y))=h(i(y'))$. But $h(i(y))=y$ and $h(i(y'))=y'$, so this means $y=y'$. Thus $i$ is one-to-one.
We now have both a one-to-one function $g:X\to Y$ and a one-to-one function $i:Y\to X$. The Schröder-Bernstein Theorem now says there exists a function $f:X\to Y$ which is both one-to-one and onto. (We can define $f$ in terms of $g$ and $i$, but it is pretty complicated--see the link above for more details).
Thus if you have one function $X\to Y$ which is one-to-one and another function $X\to Y$ which is onto, this implies $X$ and $Y$ have the same cardinality.
[As a caveat, the existence of a function $i$ as described in the second paragraph above requires the axiom of choice: for each $y\in Y$, you must choose one $x\in X$ such that $h(x)=y$. In fact, without the axiom of choice, the final result may not be true. As bof commented, there exists a one-to-one map $\mathbb{R}\to\mathbb{R}\cup\omega_1$ and also an onto map $\mathbb{R}\to\mathbb{R}\cup\omega_1$, but it is impossible to prove $\mathbb{R}$ and $\mathbb{R}\cup\omega_1$ have the same cardinality without using the axiom of choice.]
Yes. Let $f: X \to Y$ be a bijection.
Then show that $\hat{f}: \mathscr{P}(X) \to \mathscr{P}(Y)$ given by $$\hat{f}(A) := f[A] (= \{f(x): x \in A\})$$
is a bijection between their power sets.
Best Answer
By definition there is always a bijection between two sets of the same cardinality, Otherwise they wouldn't have the same cardinality.
However, calling it an "isomorphism" suggests that you're looking for a bijection that preserves some kind of additional structure, and until you tell us which kind of additional structure you want preserved, the question can't really be answered.
For most practically occurring cases, I think the answer will be "no, there exist non-isomorphic such-and-suches with the same cardinality". But there are pathological corner cases where this is not the case, for example if we consider isomorphisms between standard models of the pure predicate calculus with equality over the empty language.
Or, somewhat less trivially: For any finite field $K$, if two vector spaces over $K$ have the same cardinality, then they're isomorphic as $K$-vector spaces.