Chapter 7: Problem 3
Suppose \(\left(X, d_{X}\right),\left(Y, d_{Y}\right)\) be metric spaces and \(f: X \rightarrow Y\) is continuous. Let \(A \subset X\). a) Show that \(f(\bar{A}) \subset \overline{f(A)}\) b) Show that the subset can be proper.
Short Answer
Expert verified
a) \(f(\bar{A}) \subset \overline{f(A)}\); b) Use different functions or sets for proper subset.
Step by step solution
01
Understanding Closure and Continuous Functions
To solve (a), we need to understand what it means for a function to be continuous and what the closure of a set is in a metric space. The closure of a set \(A\), denoted \(\bar{A}\), is the set of all limit points of \(A\) along with \(A\) itself. A function \(f: X \to Y\) is continuous if for every \(x_n \to x\) in \(X\), we have \(f(x_n) \to f(x)\) in \(Y\).
02
Proving Part (a) - Continuity and Closure Relationship
Consider a point \(y \in f(\bar{A})\). By definition, \(y = f(x)\) for some \(x \in \bar{A}\). Since \(x\) is in the closure of \(A\), there exist points \(x_n \in A\) such that \(x_n \to x\). By continuity, \(f(x_n) \to f(x) = y\). Therefore, \(y\) is a limit point of \(f(A)\), and thus \(y \in \overline{f(A)}\). Hence, \(f(\bar{A}) \subset \overline{f(A)}\).
03
Proving Part (b) - Proper Subset Example
To show \(f(\bar{A})\) and \(\overline{f(A)}\) can be a proper subset, consider metric spaces \((X, d) = (\mathbb{R}, |\cdot|)\) and \((Y, d) = (\mathbb{R}, |\cdot|)\) with \(f: X \to Y\) defined by \(f(x) = x^2\). Take \(A = (-1, 1)\). Then \(\bar{A} = [-1, 1]\), and \(f(\bar{A}) = [0, 1]\). However, \(f(A) = [0, 1)\), and \(\overline{f(A)} = [0, 1]\). Therefore, \(f(\bar{A}) = \overline{f(A)} = [0, 1]\), illustrating no proper subset. For a proper example, consider \(g(x) = x^3\), where the closure equality doesn't hold in general due to different limit points.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuous Functions
In the realm of metric spaces, understanding continuous functions is vital. A function \(f: X \to Y\) is said to be continuous when for every sequence \(x_n \to x\) in \(X\), the image sequence \(f(x_n)\) converges to \(f(x)\) in \(Y\). This means that the behavior of the function on sequences in its domain should reflect consistently in its codomain.
- An essential property of continuous functions is that they maintain limits, which is why they are crucial in proving relationships between closures of sets.
- In simple terms, if you slide your input smoothly across a function without any jumps, the output should also change smoothly without disruptions.
Closure of a Set
The concept of closure in metric spaces involves expanding a set to include its limit points. The closure of a set \(A\), denoted as \(\bar{A}\), consists of all points of \(A\) plus all points that can be approached by sequences from \(A\), even if they are not in \(A\) itself.
- Limit points of a set are those that can be `touched' by points of the set through sequences. For instance, edges of an interval like \((0, 1)\) can be considered when looking at its closure \([0, 1]\).
- Formally, a point \(x\) is in the closure of \(A\) if every neighborhood around \(x\) contains a point from \(A\) other than \(x\) itself.
Limit Points
Understanding limit points helps grasp the concept of closures more deeply. A limit point of a set \(A\) in a metric space is a point where every neighborhood contains at least one point from \(A\), different from the point itself. In a practical sense, these are points where you can `approach' using points strictly in \(A\), but not necessarily in \(A\).
- An easy example is how \(0\) and \(1\) are limit points of the open interval \((0, 1)\), even though they are not included in the set itself.
- Limit points are what contribute to the `closure' of a set, reflecting how nearby points converge into exampling sets.
Subset Relationships
Subset relationships are foundational in mathematics, especially when describing the relationship between sets and their images or pre-images under functions. In the context of metric spaces and functions, understanding how the image of a closure relates to the closure of the image under continuous functions is key.
- The statement \(f(\bar{A}) \subset \overline{f(A)}\) showcases a subset relationship where the image of the closure can be smaller or equal to the closure of the image.
- However, they don't always have to be equal, as demonstrated through specific function mappings like \(g(x) = x^3\) where the output changes boundaries differently.