1. **Problem Statement:** Consider the sequence of functions $f_n:\mathbb{R} \to \mathbb{R}$ defined by $$f_n(x) = \frac{nx}{1 + n^2 x^2}.$$ We need to determine if $f_n$ converges uniformly on $\mathbb{R}$. 2. **Pointwise limit:** Fix $x \in \mathbb{R}$ and examine the limit as $n \to \infty$. For $x=0$, clearly $f_n(0)=0$ for all $n$. For $x \neq 0$: $$ \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{nx}{1 + n^2 x^2} = \lim_{n \to \infty} \frac{n x}{n^2 x^2 \left(\frac{1}{n^2 x^2} + 1\right)} = \lim_{n \to \infty} \frac{1}{n x + \frac{1}{n x}} = 0. $$ So $f_n(x) \to 0$ pointwise for all $x \in \mathbb{R}$. 3. **Check uniform convergence:** To check uniform convergence, inspect the supremum norm: $$ \sup_{x \in \mathbb{R}} |f_n(x) - 0| = \sup_{x \in \mathbb{R}} \left| \frac{nx}{1 + n^2 x^2} \right|. $$ Set $t = n x$, then: $$ |f_n(x)| = \left| \frac{nx}{1 + n^2 x^2} \right| = \frac{|t|}{1 + t^2}. $$ 4. **Find maximum of $g(t) = \frac{|t|}{1 + t^2}$:** - For $t > 0$, consider $g(t) = \frac{t}{1 + t^2}.$ - Differentiate: $$g'(t) = \frac{(1 + t^2) - t (2t)}{(1 + t^2)^2} = \frac{1 + t^2 - 2 t^2}{(1 + t^2)^2} = \frac{1 - t^2}{(1 + t^2)^2}.$$ - Critical points at $t = 1$ and $t = -1$. - At $t=1$, $$g(1) = \frac{1}{1 + 1} = \frac{1}{2}.$$ Similarly for negative $t$, the same maximum value $\frac{1}{2}$ occurs. So $$ \sup_{x \in \mathbb{R}} |f_n(x)| = \frac{1}{2} \quad \text{for all } n. $$ 5. **Conclusion about uniform convergence:** Since the supremum does not go to zero, the convergence is not uniform on $\mathbb{R}$. --- 6. **(a) Schröder-Bernstein Theorem Statement:** If there exist injective functions $f : A \to B$ and $g : B \to A$ between sets $A$ and $B$, then there exists a bijection $h : A \to B$. **Proof outline:** - Define subsets of $A$ based on the iterative image and preimage under $f$ and $g$. - Construct $h$ piecewise by carefully matching elements. - The details ensure $h$ is bijective. (Full proof is classical and can be found in set theory textbooks.) --- 7. **(b) In a metric space $(M,d)$, show that $M$ and $\emptyset$ are both open and closed:** - By definition, the empty set and the whole set are open. - Their complements are respectively $M$ and $\emptyset$, which are open. - A set whose complement is open is closed. - Therefore, both $M$ and $\emptyset$ are closed and open. --- **Summary:** - $f_n$ does **not** converge uniformly on $\mathbb{R}$. - Schröder-Bernstein Theorem establishes bijection from injections. - $M$ and $\emptyset$ are always open and closed in any metric space.