To tackle these questions, let's carefully analyze each part and use properties of uniformly continuous functions to derive the desired results.

(a) Existence of $a \in (0, b]$

Proof:

1. Since $f$ is uniformly continuous on $(0, b]$, for $\epsilon = 1$, there exists $\delta > 0$ such that for all $x, y \in (0, b]$, if $|x - y| < \delta$, then $|f(x) - f(y)| < 1$.

2. Choose $a = \min\{b, \delta\}$. This ensures that $a$ lies in $(0, b]$.

3. Now, for all $x, y \in (0, a]$ with $|x - y| < \delta$ (and since $a \leq \delta$), we have $|f(x) - f(y)| < 1$.

Conclusion: Thus, there exists an $a \in (0, b]$ such that $|f(x) - f(y)| < 1$ for all $x, y \in (0, a]$.

Bounding $f$ on $(0, a]$:

1. Choose any $x_0 \in (0, a]$.

2. Since $|f(x) - f(x_0)| < 1$ for all $x \in (0, a]$, it follows that: $|f(x)| \leq |f(x_0)| + 1 \text{ for all } x \in (0, a]$

3. Let $K = |f(x_0)| + 1$. Then $K > 0$ and $|f(x)| \leq K$ for all $x \in (0, a]$.

(b) Boundedness of $f$ on $(0, b]$

Proof:

1. Divide $(0, b]$ into two parts: $(0, a]$ and $[a, b]$.

2. From part (a), $f$ is bounded on $(0, a]$ by $K$.

3. On the closed interval $[a, b]$, the function $f$ is uniformly continuous (because $f$ is uniformly continuous on the whole $(0, b]$) and continuous on a compact set, hence bounded.

4. Let $M = \max_{x \in [a, b]} |f(x)|$.

5. Therefore, $f$ is bounded on $(0, b]$ by $\max(K, M)$.

(c) Uniform continuity of $g(x) = \frac{1}{x} \sin\left(\frac{1}{x}\right)$ on $(0, 1]$

Reasoning:

1. Let's analyze the behavior of $g(x)$ as $x \to 0$.

2. Consider any two points $x, y \in (0, 1]$.

3. If $|x - y| < \delta$, where $\delta$ is small, the behavior of $g$ depends on $\frac{1}{x}$ and $\frac{1}{y}$. For small $x$, $\frac{1}{x}$ becomes large, causing the sine function to oscillate rapidly.

4. This rapid oscillation implies that for small changes in $x$, $|g(x) - g(y)|$ can be large due to the factor $\frac{1}{x}$.

5. Hence, there does not exist a single $\delta$ that works uniformly for all $\epsilon > 0$ for small $x$, implying $g(x)$ is not uniformly continuous on $(0, 1]$.

Conclusion: The function $g(x) = \frac{1}{x} \sin\left(\frac{1}{x}\right)$ is not uniformly continuous on $(0, 1]$.

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Do you want details or have any questions?

Here are 8 related questions for further exploration:

1. What are some other examples of functions that are uniformly continuous on a given interval?
2. How does the definition of uniform continuity differ from regular continuity?
3. Can you provide an example of a function that is continuous but not uniformly continuous?
4. How can one prove that a function is not uniformly continuous on a given interval?
5. What is the importance of bounding a function in real analysis?
6. How does one handle uniform continuity on half-open intervals compared to closed intervals?
7. Are there any special properties of uniformly continuous functions on bounded intervals?
8. How does the uniform continuity of a function affect its integrability?

Tip: When proving properties about uniformly continuous functions, consider dividing the interval into manageable subintervals, especially when dealing with unbounded or partially bounded domains.