The intuition behind conditional probability and independence in the case of different sample space I came up with this question when doing this problem: 

In throwing a pair of dice, let A be the event that "the first die turns up odd", B the event that "the second die turns up odd", and C the event that "the total number of spots is odd". Then since P(C|A)=P(C), we proved that A and C are independent (B and C are similar).

Normally, we can use Venn diagram to explain conditional probability. And the intuition for P(X|Y) is we can first denote the number of outcomes for XY and Y as n(XY) and n(Y) (graphically, the corresponding area), then both divided by the number of trials n, we can finally use (n(XY)/n)/(n(Y)/n) to approximate "in n trials, the probability that X happens given Y happens." Note that intuitively, we believe relative frequency converges to probability.  
But in our case, the thing gets a bit complicated. The sample space for C is {2,3,...,12} and the sample space for A is {1,...,6} , they are different sample space! We can't explain in the traditional way of doing an experiment and analyzing the intersection of events. So I wonder the true intuition behind the definition of conditional probability (P(X|Y)=P(XY)/P(Y)), and the definition of independence (P(XY)=P(X)P(Y) or P(X|Y)=P(X))
FYI, try compare with a normal question like this one: Let A be the event that a card picked at random from a full deck is a spade, and B the event that it is a queen. Then A and B are independent.
 A: The sample space here is simply all possible rolls of two (distinguishable) dice.
There are various random functions on the sample space: e.g. the function


*

*The value of the first die

*The value of the second die

*The sum of the two values


Your mistake is confusing the ranges of these random functions as being sample spaces.
While "the number is odd" is a proposition one can ask about integers, that is not what event $C$ is. Event $C$ is the application of that proposition to the value of the third random variable mentioned above, and is still an event on the sample space of all outcomes of two die rolls.
For example, event $B$ is true on the outcomes
$$\{ (1,1), (2,1), (3,1), (4,1), (5,1), (6,1), 
\\ (1,3), (2,3), (3,3), (4,3), (5,3), (6,3),
\\ (1,5), (2,5), (3,5), (4,5), (5,5), (6,5) \}$$

Usually, all of the random variables and events you consider at one time will all be defined on the same sample space.
In those circumstances where you had considered two separate problems, and then for some reason decided to consider them both again at the same time, you would combine the sample spaces together by taking the Cartesian product.
e.g. if you studied the first die roll on its own with sample space $\{1,2,3,4,5,6\}$, then studied the second die roll on its own with sample $\{1,2,3,4,5,6\}$, and then decided to study them both together, you'd switch everything over to the sample space consisting of all pairs of rolls.
e.g. if you had an event $E=\{2,3\}$ when studying the first die roll, when you switch over to the combined sample space, you would extend the event by just considering the first coordinate: the extension of $E$ would be $\{(2,1), (2,2), \ldots, (2,6), (3,1), \ldots, (3,6) \}$.
