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A set $S$ is called star shaped if there exist a point $z_0\in S$ such that the line segment between $z_0$ and any point in $S$ is contained in $S$. Prove that a star shaped set is simply connected.

Let $\gamma_1$ and $\gamma_2$ be two closed curves with the same initial and end points $z_0\in S$ which are both parameterized on $I$.

Define

$$

F:I^2\to S \quad \text{by} \quad

F(s,t) = (1 - t)\gamma_1(s) + t\gamma_2(s),

$$

where $I = [0,1]$.

Then $F(s,t)$ is uniformly continuous since $I^2$ is compact and $F(s,t)$ is the sum of continuous functions.

So $F(s,0) = \gamma_1(s)$, $F(s,1) = \gamma_2(s)$, $F(0,t) = z_0$, and $F(1,t) = z_0$.

Additionally, any closed $\gamma_i$ is homotopic to a point in $S$.

So $\int_{\gamma_i}F = 0$.

Therefore $S$ is simply connected.

Correct?

Let $\gamma_1$ and $\gamma_2$ be two closed curves with the same initial and end points $z_0\in S$ which are both parameterized on $I$.

Define

$$

F:I^2\to S \quad \text{by} \quad

F(s,t) = (1 - t)\gamma_1(s) + t\gamma_2(s),

$$

where $I = [0,1]$.

Then $F(s,t)$ is uniformly continuous since $I^2$ is compact and $F(s,t)$ is the sum of continuous functions.

So $F(s,0) = \gamma_1(s)$, $F(s,1) = \gamma_2(s)$, $F(0,t) = z_0$, and $F(1,t) = z_0$.

Additionally, any closed $\gamma_i$ is homotopic to a point in $S$.

So $\int_{\gamma_i}F = 0$.

Therefore $S$ is simply connected.

Correct?

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