Prove that the following sets are connected in R2: (a) {(x,y) ∈ R2 : x2 +y2 = 1}. (b) {(x,y) ∈ R2 : x2 a2 + y2 b2 = 1}. (c) {(x,y) ∈ R2 : y = x2}.

To prove that the given sets in \( \mathbb{R}^2 \) are connected, we will use the following key fact about connectedness:

### Fact about Connected Sets:
- A subset \( A \subseteq \mathbb{R}^2 \) is **connected** if it cannot be written as the union of two non-empty, disjoint, open sets in \( A \). Intuitively, a connected set is "in one piece" without any gaps.
- For continuous images of connected sets, the image remains connected. For example, continuous transformations (like a parametrization or a curve) of intervals or connected sets in \( \mathbb{R} \) will yield connected sets in \( \mathbb{R}^2 \).

Now we will prove the connectedness of each set.

### (a) The unit circle: \( \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 = 1\} \)

This is the equation of the unit circle centered at the origin with radius 1.

#### Proof:
- The unit circle can be parametrized by a continuous map \( f: [0, 2\pi] \to \mathbb{R}^2 \) as follows:

\[
f(\theta) = (\cos \theta, \sin \theta), \quad \theta \in [0, 2\pi].
\]

- The interval \([0, 2\pi]\) is connected in \( \mathbb{R} \), and since the image of a connected set under a continuous map is also connected, the image of \( f \), which is \( \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 = 1\} \), is connected in \( \mathbb{R}^2 \).

Thus, the unit circle is connected.

### (b) The ellipse: \( \{(x, y) \in \mathbb{R}^2 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\} \)

This is the equation of an ellipse centered at the origin with semi-major axis \( a \) along the \( x \)-axis and semi-minor axis \( b \) along the \( y \)-axis.

#### Proof:
- The ellipse can be parametrized by a continuous map \( g: [0, 2\pi] \to \mathbb{R}^2 \) as follows:

\[
g(\theta) = (a \cos \theta, b \sin \theta), \quad \theta \in [0, 2\pi].
\]

- Again, since \( [0, 2\pi] \) is connected and \( g \) is continuous, the image of \( g \), which is \( \{(x, y) \in \mathbb{R}^2 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\} \), is connected.

Thus, the ellipse is connected.

### (c) The parabola: \( \{(x, y) \in \mathbb{R}^2 : y = x^2\} \)

This is the equation of a parabola opening upward along the \( y \)-axis.

#### Proof:
- The parabola can be parametrized by the continuous map \( h: \mathbb{R} \to \mathbb{R}^2 \) as follows:

\[
h(t) = (t, t^2), \quad t \in \mathbb{R}.
\]

- The real line \( \mathbb{R} \) is connected, and since the map \( h \) is continuous, the image of \( h \), which is \( \{(x, y) \in \mathbb{R}^2 : y = x^2\} \), is connected.

Thus, the parabola is connected.

### Conclusion:
We have shown that the following sets are connected:
1. \( \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 = 1\} \) (the unit circle),
2. \( \{(x, y) \in \mathbb{R}^2 : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\} \) (the ellipse),
3. \( \{(x, y) \in \mathbb{R}^2 : y = x^2\} \) (the parabola).

Would you like more details on any part of the proof?

### 5 Related Questions:
1. Can a set be disconnected if its continuous parametrization is not injective?
2. How can you use continuity to prove that more general shapes (like hyperbolas) are connected?
3. What are the implications of compactness for connected sets in \( \mathbb{R}^2 \)?
4. Can a set with a hole, such as a circle with a missing point, be connected?
5. How does the connectedness of curves change when they are defined piecewise?

### Tip:
When working with connected sets, parametrizations are a powerful tool. They allow you to leverage the connectedness of simple intervals or domains to prove connectedness in higher-dimensional spaces.

Explore how to prove the connectedness of three key sets in R2: the unit circle, an ellipse, and a parabola. This detailed mathematical proof covers concepts from topology, parametric equations, and continuous maps, suitable for university-level students.

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