Problem 5
Question
1-27. Prove that \(\left\\{x \in \mathbf{R}^{n}:|x-a|
Step-by-Step Solution
Verified Answer
The set is open because it is the preimage of an open set under a continuous function.
1Step 1 - Understand the definition of an open set
A set is defined as open if for every point within the set, there is a neighborhood around that point which is entirely contained within the set.
2Step 2 - Define the set and function
We need to prove that \(\left\{x \in \mathbf{R}^{n}:|x-a|
3Step 3 - Show the continuity of the function
The function \(f(x)=|x-a|\) is continuous because the norm operation and subtraction are both continuous operations in \(\mathbf{R}^{n}\).
4Step 4 - Use the preimage of an open set
The set \(f^{-1}((-fty,r))\) is the preimage of the open interval \((-fty, r) \) under the continuous function \(f\). The preimage of an open set under a continuous function is also open.
5Step 5 - Conclude that the original set is open
Since \(f\) is continuous and \((-fty, r)\) is open, \(f^{-1}((-fty,r))\) is also open. Thus, \(\left\{x \in \mathbf{R}^{n}:|x-a|
Key Concepts
Open SetsContinuous FunctionsPreimages in TopologyMetric Spaces
Open Sets
In topology, an open set is a fundamental concept.
An open set is defined as a set where, for every point within it, there exists a neighborhood entirely contained within that set.
In simpler terms, you can imagine it as a set where you can always draw a small 'bubble' around any point in it without stepping outside the set.
For instance, in the set \(\{x \in \mathbf{R}^{n}:|x-a| This concept helps us understand and classify different spaces and functions according to their characteristics in topology.
An open set is defined as a set where, for every point within it, there exists a neighborhood entirely contained within that set.
In simpler terms, you can imagine it as a set where you can always draw a small 'bubble' around any point in it without stepping outside the set.
For instance, in the set \(\{x \in \mathbf{R}^{n}:|x-a|
Continuous Functions
A function is continuous if small changes in the input result in small changes in the output.
Specifically, for any function \(f: X \rightarrow Y\), if whenever \(x \rightarrow a\) then \(f(x) \rightarrow f(a),\) the function is continuous.
In our specific case, the function \(f(x)=|x-a|\) maps each point in \(\mathbf{R}^{n}\) to a real number representing its distance from \(a.\) Distance is inherently a continuous measurement because minor adjustments in \(x\)'s value will result in similarly minor adjustments in distance.
This continuity is crucial for analyzing and describing the behavior of functions within different spaces, ensuring smooth transitions and lack of sudden jumps.
Specifically, for any function \(f: X \rightarrow Y\), if whenever \(x \rightarrow a\) then \(f(x) \rightarrow f(a),\) the function is continuous.
In our specific case, the function \(f(x)=|x-a|\) maps each point in \(\mathbf{R}^{n}\) to a real number representing its distance from \(a.\) Distance is inherently a continuous measurement because minor adjustments in \(x\)'s value will result in similarly minor adjustments in distance.
This continuity is crucial for analyzing and describing the behavior of functions within different spaces, ensuring smooth transitions and lack of sudden jumps.
Preimages in Topology
A preimage refers to the set of all points that map to a given set under a function.
In other words, if you have a function \(f: X \rightarrow Y\) and a subset \(B \subset Y\), the preimage of \(B\) under \(f\) is the set of all points in \(X\) that \(f\) maps into \(B.\)
Mathematically, it is represented as: \[ f^{-1}(B) = \{ x \in X \mid f(x) \in B\} \].
For instance, with \(f: \mathbf{R}^{n} \rightarrow \mathbf{R}\) and \(f(x)=|x-a|\), if we consider an open interval \((-\infty, r)\), the preimage is the set of all \(x\) in \(\mathbf{R}^{n}\) with \(f(x) < r\).
This is particularly significant because the preimage of an open set under a continuous function is always open.
In other words, if you have a function \(f: X \rightarrow Y\) and a subset \(B \subset Y\), the preimage of \(B\) under \(f\) is the set of all points in \(X\) that \(f\) maps into \(B.\)
Mathematically, it is represented as: \[ f^{-1}(B) = \{ x \in X \mid f(x) \in B\} \].
For instance, with \(f: \mathbf{R}^{n} \rightarrow \mathbf{R}\) and \(f(x)=|x-a|\), if we consider an open interval \((-\infty, r)\), the preimage is the set of all \(x\) in \(\mathbf{R}^{n}\) with \(f(x) < r\).
This is particularly significant because the preimage of an open set under a continuous function is always open.
Metric Spaces
Metric spaces provide a framework for discussing distances between points.
In a metric space, we use a function called a metric to quantify how far apart points are.
Our focus is primarily on \( \mathbf{R}^{n} \) (Euclidean space) where the distance between any two points is defined by the Euclidean norm \(|x-a|\).
This metric allows us to define open sets using balls: the set \(\left\{ x \in \mathbf{R}^{n}: |x-a| These open balls form the basis for open sets in metric spaces, which in turn helps in understanding the topology of these spaces.
In a metric space, we use a function called a metric to quantify how far apart points are.
Our focus is primarily on \( \mathbf{R}^{n} \) (Euclidean space) where the distance between any two points is defined by the Euclidean norm \(|x-a|\).
This metric allows us to define open sets using balls: the set \(\left\{ x \in \mathbf{R}^{n}: |x-a|
Other exercises in this chapter
Problem 4
1-17. Construct a set \(A \subset[0,1] \times[0,1]\) such that \(A\) contains at most one point on each horizontal and each vertical line but boundary \(A=[0,1]
View solution Problem 4
Prove that ||\(x|-| y|| \leq|x-y|\).
View solution Problem 5
1-17. Construct a set \(A \subset[0,1] \times[0,1]\) such that \(A\) contains at most one point on each horizontal and each vertical line but boundary \(A=[0,1]
View solution Problem 6
1-28. If \(A \subset \mathbf{R}^{n}\) is not closed, show that there is a continuous function \(f: A \rightarrow \mathbf{R}\) which is unbounded. Hint: If \(x \
View solution