Solution to Question 3
Solution to Question 3
Part (a)
Problem Statement:
If \(X\) is a connected topological space and \(f: X \to Y\) is a continuous and surjective function, then \(Y\) is connected.
Proof:
We need to prove that \(Y\) is a connected topological space. We will use the definition of connectedness by contradiction. A topological space is connected if it cannot be written as the union of two non-empty, disjoint open sets.
Assume, for the sake of contradiction, that \(Y\) is not connected. This means there exist two non-empty, disjoint open sets \(U\) and \(V\) in \(Y\) such that \(Y = U \cup V\).
Since \(f\) is a continuous function, the pre-images of open sets are open. Therefore, \(f^{-1}(U)\) and \(f^{-1}(V)\) are open sets in \(X\).
Given that \(f\) is a surjective function, for every \(y \in Y\), there exists an \(x \in X\) such that \(f(x) = y\).
Since \(Y = U \cup V\), we can write \(X\) as the union of the pre-images: \(X = f^{-1}(Y) = f^{-1}(U \cup V) = f^{-1}(U) \cup f^{-1}(V)\).
Since \(U\) and \(V\) are non-empty, their pre-images \(f^{-1}(U)\) and \(f^{-1}(V)\) are also non-empty. This is because \(f\) is surjective. If \(f^{-1}(U)\) were empty, then \(U\) would be the image of the empty set under \(f\), which is empty, contradicting the assumption that \(U\) is non-empty. The same logic applies to \(V\).
Since \(U \cap V = \emptyset\), we have \(f^{-1}(U \cap V) = f^{-1}(\emptyset) = \emptyset\).
Also, \(f^{-1}(U \cap V) = f^{-1}(U) \cap f^{-1}(V)\).
Thus, \(f^{-1}(U) \cap f^{-1}(V) = \emptyset\).
So, we have partitioned \(X\) into two non-empty, disjoint open sets, \(f^{-1}(U)\) and \(f^{-1}(V)\). This means that \(X\) is not connected.
This contradicts our initial assumption that \(X\) is a connected topological space. Therefore, our assumption that \(Y\) is not connected must be false.
Hence, \(Y\) must be connected.
Part (b)
Problem Statement:
Find \(Fr(A)\), where \(A=(0,1)\), a subset of the topological space \((\mathbb{R}, d)\), where \(d\) is the standard metric topology on \(\mathbb{R}\).
Solution:
The standard topology on \(\mathbb{R}\) is the one induced by the standard metric, where open sets are unions of open intervals.
The closure of a set \(A\), denoted \(\bar{A}\), is the smallest closed set containing \(A\). In \(\mathbb{R}\) with the standard topology, the closure of an open interval \((a,b)\) is the closed interval \([a,b]\).
Thus, for the set \(A = (0,1)\), its closure is \(\bar{A} = [0,1]\).
The interior of a set \(A\), denoted \(A^{\circ}\), is the largest open set contained in \(A\). For the set \(A = (0,1)\), the interior is the set itself, as it is an open interval.
Thus, \(A^{\circ} = (0,1)\).
The frontier (or boundary) of a set \(A\), denoted \(Fr(A)\) or \(\partial A\), is defined as the difference between the closure and the interior of the set.
\(Fr(A) = \bar{A} \setminus A^{\circ}\).
Using the results from above, we have: \(Fr(A) = [0,1] \setminus (0,1)\).
The set difference removes all points in the open interval \((0,1)\) from the closed interval \([0,1]\). The only points remaining are the endpoints.
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