Show That A 1 ∪ A 2 A_{1}\cup A_{2} A 1 ​ ∪ A 2 ​ Is Not Connected By Path.

Introduction

In real analysis, a connected set is a set that cannot be represented as the union of two or more disjoint non-empty open sets. In this article, we will show that the set $A = A_{1} \cup A_{2}$ is not connected by path, where $A_{1}$ and $A_{2}$ are subsets of $\mathbb{R}^{2}$.

Definition of Connected Sets

A set $A$ is said to be connected if it cannot be written as the union of two or more disjoint non-empty open sets. In other words, a set is connected if it is not possible to separate it into two or more disjoint open sets.

Definition of Path Connected Sets

A set $A$ is said to be path connected if for any two points $x, y \in A$, there exists a continuous function $f: [0,1] \to A$ such that $f(0) = x$ and $f(1) = y$. In other words, a set is path connected if it is possible to draw a continuous curve between any two points in the set.

The Set $A_{1}$

The set $A_{1}$ is defined as:

$$A_{1} = \{\big(x, \sin\big(\frac{1}{x}\big)\big); x \in \big(0, \frac{2}{\pi}\big]\}$$

This set consists of points in the plane whose x-coordinate is in the interval $(0, \frac{2}{\pi}]$ and whose y-coordinate is the sine of the reciprocal of the x-coordinate.

The Set $A_{2}$

The set $A_{2}$ is defined as:

$$A_{2} = \{0\} \times [-1,1]$$

This set consists of points in the plane whose x-coordinate is 0 and whose y-coordinate is in the interval $[-1,1]$.

The Set $A = A_{1} \cup A_{2}$

The set $A$ is the union of the sets $A_{1}$ and $A_{2}$. It consists of all points in the plane that are either in $A_{1}$ or in $A_{2}$.

Showing that $A$ is not Connected by Path

To show that $A$ is not connected by path, we need to show that there exist two points in $A$ that cannot be joined by a continuous curve.

Let $x_{1} = (0, 1) \in A_{2}$ and $x_{2} = (\frac{2}{\pi}, 0) \in A_{1}$. We will show that there is no continuous curve that joins $x_{1}$ and $x_{2}$.

Assume that there is a continuous curve that joins $x_{1}$ and $x_{2}$

Let $f: [0,1] \to A$ be a continuous curve that joins $x_{1}$ and $x_{2}$. Then, $f(0) = x_{1} = (0, 1)$ and $f(1) = x_{2} = (\frac{2}{\pi}, 0)$.

Since $f$ is continuous, the image of $f$ is a connected set. Therefore, the image of $f$ must be either in $A_{1}$ or in $A_{2}$.

Case 1: The image of $f$ is in $A_{1}$

If the image of $f$ is in $A_{1}$, then for all $t \in [0,1]$, we have $f(t) = (x, \sin\big(\frac{1}{x}\big))$ for some $x \in (0, \frac{2}{\pi}]$.

Since $f(0) = (0, 1)$, we have $\sin\big(\frac{1}{0}\big) = 1$. However, this is a contradiction since $\sin\big(\frac{1}{0}\big)$ is not defined.

Case 2: The image of $f$ is in $A_{2}$

If the image of $f$ is in $A_{2}$, then for all $t \in [0,1]$, we have $f(t) = (0, y)$ for some $y \in [-1,1]$.

Since $f(1) = (\frac{2}{\pi}, 0)$, we have $y = 0$. However, this is a contradiction since $y$ cannot be equal to 0 for all $t \in [0,1]$.

Conclusion

We have shown that there is no continuous curve that joins $x_{1}$ and $x_{2}$. Therefore, the set $A = A_{1} \cup A_{2}$ is not connected by path.

Conclusion

Q: What is a connected set?

A: A set $A$ is said to be connected if it cannot be written as the union of two or more disjoint non-empty open sets. In other words, a set is connected if it is not possible to separate it into two or more disjoint open sets.

Q: What is a path connected set?

A: A set $A$ is said to be path connected if for any two points $x, y \in A$, there exists a continuous function $f: [0,1] \to A$ such that $f(0) = x$ and $f(1) = y$. In other words, a set is path connected if it is possible to draw a continuous curve between any two points in the set.

Q: How do you show that a set is not connected by path?

A: To show that a set is not connected by path, you need to show that there exist two points in the set that cannot be joined by a continuous curve. This can be done by assuming that there is a continuous curve that joins the two points and then showing that this assumption leads to a contradiction.

Q: What is an example of a set that is not connected by path?

A: The set $A = A_{1} \cup A_{2}$, where $A_{1} = \{\big(x, \sin\big(\frac{1}{x}\big)\big); x \in \big(0, \frac{2}{\pi}\big]\}$ and $A_{2} = \{0\} \times [-1,1]$, is an example of a set that is not connected by path. We showed in the previous article that there exist two points in $A$ that cannot be joined by a continuous curve.

Q: What is the difference between a connected set and a path connected set?

A: A connected set is a set that cannot be written as the union of two or more disjoint non-empty open sets. A path connected set is a set that is possible to draw a continuous curve between any two points in the set. While all path connected sets are connected, not all connected sets are path connected.

Q: Can a set be both connected and path connected?

A: Yes, a set can be both connected and path connected. For example, the set $\mathbb{R}$ is both connected and path connected.

Q: Can a set be connected but not path connected?

A: Yes, a set can be connected but not path connected. For example, the set $A = A_{1} \cup A_{2}$, where $A_{1} = \{\big(x, \sin\big(\frac{1}{x}\big)\big); x \in \big(0, \frac{2}{\pi}\big]\}$ and $A_{2} = \{0\} \times [-1,1]$, is connected but not path connected.

Q: Can a set be path connected but not connected?

A: No, a set cannot be path connected but not connected. If a set is path connected, then it is also connected.

Conclusion

In this article, we have answered some common questions about connected sets and path connected sets. We have discussed the definitions of connected sets and path connected sets, and we have provided examples of sets that are connected but not path connected and sets that are path connected but not connected.