Continuous functions on Munkres's book

[bigli]

This is not a homework but it is a question in my mind.please guide me.

Let X and Y be topological spaces,let f : X -----> Y is a function.

when the following statements are equivalent?:

1) f is continuous

2) f(A') is subset of f(A)' ,for every A subset of X.

Symbols: A' i.e limit points set of A ,and f(A)' i.e limit points set of f(A).

pointing out: look to theorem 18-1 (page 104) from Munkres's book (TOPOLOGY 2edition 2000) and exercise 2 (page 111) from Munkres's book.

 

[morphism]

(2) will always imply (1). But (1) doesn't necessarily imply (2), as you probably already know. So we might try looking for extra conditions to impose on f.

Let's suppose X and Y are arbitrary topological spaces, and f:X->Y is an arbitrary continuous function. Let A be some subset of X, and let x be in A'. We want to show that f(A') $\subset$ f(A)', so we want f(x) to be in f(A)'. If x sits in A, then f(x) will sit in f(A). So it would be necessary to have that f(A) $\cap$ f(A)' $\neq \emptyset$. But this is not always true. So it appears that in order to say anything intelligent, we would have to impose some conditions on the nature of A (or X or Y), or on the topologies involved.

 

[tacman]

Firstly, it is important to understand the definition of continuity for a function between two topological spaces. According to Munkres's book, a function f : X -----> Y is continuous if for every open set V in Y, the pre-image f^-1(V) is open in X. This means that the inverse image of an open set in Y is an open set in X.

Now, let us consider the two statements given in the question.

1) f is continuous

2) f(A') is subset of f(A)' ,for every A subset of X.

To show that these two statements are equivalent, we need to prove that one implies the other and vice versa.

Firstly, let us assume that f is continuous. This means that for any open set V in Y, the pre-image f^-1(V) is open in X. Now, let A be a subset of X. We want to show that f(A') is a subset of f(A)'.

Let y be a limit point of f(A). This means that for any open set V containing y, there exists a point x in A such that f(x) is in V. Since f is continuous, the pre-image f^-1(V) is open in X. This means that there exists an open set U in X such that x is in U and f(U) is a subset of V.

Now, since x is a limit point of A, there exists a point a in A such that a is in U. This means that f(a) is in f(U), which is a subset of V. Therefore, y is a limit point of f(A).

Hence, we have shown that f(A') is a subset of f(A)'.

Conversely, let us assume that f(A') is a subset of f(A)' for every subset A of X. We want to show that f is continuous.

Let V be an open set in Y. We want to show that the pre-image f^-1(V) is open in X. Let x be a point in f^-1(V). This means that f(x) is in V. Since V is open, there exists a point y in V such that y is a limit point of V.

Now, since f(A') is a subset of f(A)' for every subset A of X, we know that y is also a limit point of