How Many Continuous and Open Mappings Exist Between Discrete Topological Spaces?

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The discussion focuses on the number of continuous and open mappings between discrete topological spaces defined by the sets X = {1, 2, 3, 4} and Y = {1, 2, 3}. Initially, the user proposed that the number of mappings is calculated using the formula \(\sum_{i = 1}^3 \frac{3!}{i!}\), but later corrected this to 81, derived from the concept of 4-th class variations of 3 elements with repetition. The identity mapping f(x) = x is questioned regarding its continuity and homeomorphism status, particularly due to the element 4 in X not having a corresponding element in Y.

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Another problem whose answer I'd like to check. Thanks in advance.

Let X = {1, 2, 3, 4} and Y = {1, 2, 3}. Let P(X) and P(Y) be the power sets of X and Y, respectively.

i) How many continuous mappings are there from the discrete topological spaces (X, P(X)) to (Y, P(Y))?

Well, I figured that every mapping we can define between these topologies is open, since for any open set in Y (i.e. any power set), the preimage must again be a power set in X, so the total number would be \sum_{i = 1}^3 \frac{3!}{i!}.

ii) How many open mappings are there?

The same answer.

iii) Is the identity mapping f(x) = x continuous, and if so, is it a homeomorphism?

Here I'm a bit confused, since we can't map 4 to 4, since 4 is not an element of Y. Shouldn't all the elements of the domain X be mapped into some element of Y?
 
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Just a second, I found an error in i) and ii), will review and post a bit later.
 
A correction of my answer to i) and ii) - the number would be the number of 4-th class variations of 3 elements with repetition, i.e. n = 3^4 = 81.
 

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