Prove that two sequences converge or are Cauchy

Question

I have the two following implications to prove. The first is this:

Assume that $\forall k\in \mathbb{N}, \{f_n(k)\}_{n=1}^\infty \subset \mathbb{R}$ converges in $\mathbb{R}$. Then prove that $\{f_n\}_{n=1}^\infty \subset \mathbb{R}^\mathbb{N}$ is convergent in the metric space $(\mathbb{R}^\mathbb{N}, d)$. I have already proven that d is a metric; it is defined by: $$ d(f,g)= \sum_{k=1}^\infty 2^{-k} \min\{1, |f(k)-g(k)| \} .$$

For this implication, I have the following so far: Assume that $\forall k\in \mathbb{N}, \{f_n(k)\}_{n=1}^\infty \subset \mathbb{R}$ converges in $\mathbb{R}$ to some point $p\in \mathbb{R}$. Then, for all $\epsilon>0$, there exists some integer $N$ such that $n\geq N\rightarrow d(f_n(k), p)=|f_n(k)-p|< \epsilon$. Note that $|f_n(k)-p|\leq \min\{1,|f_n(k)-p|\}$ and $0<2^{k}<1$ for all $1\leq k< \infty$, so then for the same set of $k$ values we have that $2^{-k} \min\{1,|f_n(k)-p|\}< |f_n(k)-p|<\epsilon$.

From here, I wanted to add all the $2^{-k} \min\{1,|f_n(k)-p|\}$ was less that $\epsilon \,\times $ the number of $k$'s, but there are infinite $k$ values so that wouldn't really work. Also, I realized that I assumed all the sequences for the different $k$ values converge to the same point, which I don't think is guaranteed. Would anyone be able to point me in the right direction or validate if anything I've shown is very useful?

For the second implication, I need to show that given that $\{f_n\}_{n=1}^\infty \subset \mathbb{R}^\mathbb{N}$ converges in the metric space $(\mathbb{R}^\mathbb{N}, d)$, show that $\forall k\in \mathbb{N}, \{f_n(k)\}_{n=1}^\infty \subset \mathbb{R}$ is a Cauchy sequence in $\mathbb{R}$. I believe this will be similar to the first implication I need to prove, but I'm not too sure on how to start.

Answer 1

Define $f(k)=\lim_{n\rightarrow\infty}f_{n}(k)$. Given $\epsilon>0$, find some positive integer $N$ such that $\displaystyle\sum_{k\geq N+1}\dfrac{1}{2^{k}}<\epsilon$. We can find a positive integer $M$ such that $|f_{n}(k)-f(k)|<\epsilon/N$ for all $n\geq M$, $k=1,...,N$. Now for all $n\geq M$, \begin{align*} d(f_{n},f)&=\sum_{k=1}^{\infty}2^{-k}\min\{1,|f_{n}(k)-f(k)|\}\\ &=\sum_{k=1}^{N}2^{-k}\min\{1,|f_{n}(k)-f(k)|\}+\sum_{k\geq N+1}2^{-k}\min\{1,|f_{n}(k)-f(k)|\}\\ &\leq\sum_{k=1}^{N}|f_{n}(k)-f(k)|+\sum_{k\geq N+1}2^{-k}\\ &<\sum_{k=1}^{N}\dfrac{\epsilon}{N}+\epsilon\\ &=2\epsilon. \end{align*}

For the second mission, let $f_{n}\rightarrow f$ in $d$ for some $f$. Fix a positive integer $k$ and let $\epsilon\in(0,1)$. Consider the number $\epsilon/2^{k}$, then for some positive integer $N$ we have $d(f_{n},f)<\epsilon/2^{k}$. If $|f_{n}(k)-f(k)|\geq 1$, then $2^{-k}=2^{-k}\min\{1,|f_{n}(k)-f(k)|\}\leq d(f_{n},f)<\epsilon/2^{k}<2^{-k}$, a contradiction. So we must have $\min\{1,|f_{n}(k)-f(k)|\}=|f_{n}(k)-f(k)|$, and hence $2^{-k}|f_{n}(k)-f(k)|<\epsilon/2^{k}$ for all such $n$. In other words, $|f_{n}(k)-f(k)|<\epsilon$ for all $n\geq N$, so $\{f_{n}(k)\}_{n=1}^{\infty}$ is convergent.

Answer 2

I'll prove the first proposition for you.

For each $k \in \mathbb{N}$, $f_n(k)$ converges to some $f(k)$. Let $f(.)$ be the function to which the $f_n(.)$ converge point-wise.

Let $\epsilon > 0$. You need to show that there is $N$ such that for all $n \ge N$ $$ d(f_n,f) \le \epsilon $$ Notice that there exists $K$ such that $$ \sum_{k=K+1}^\infty 2^{-k} \le \epsilon/2. $$ Since the $f_n(k)$ converges to $f(k)$, for $k \in \mathbb{N}$ you can pick $N_k$ such that for all $n \ge N_k$ $$ |f_n(k)-f(k)| \le \epsilon/2 $$ Let $N = \max\{N_1, ..., N_K\}$. Then for $n \ge N$ $$ d(f_n,f) = \sum_{k=1}^K 2^{-k}min\{|f_n((k)-f(k)|,1\}+\sum_{k=K+1}^\infty 2^{-k} \le \epsilon/2 + \epsilon/2 = \epsilon $$ This proof is an introduction into a class of proofs that uses two or three $\epsilon$'s to prove convergence in a space of functions. The Cauchy convergence will work in a similar way.

Answer 3

For a more conceptual proof, notice that $d$ induces the product topology on $\mathbb{R}^\mathbb{N}$:

Let $f \in \mathbb{R}^\mathbb{N}$ and let $B_d(f, r)$ be an open ball with respect to the metric $d$. We'll show that there exists a basis element of the product topology containing $f$ and contained in $B_d(f, r)$.

Indeed, let $n_0 \in \mathbb{N}$ be such that $\sum_{n=n_0+1}^\infty \frac1{2^n} < \frac{r}2$ and consider:

$$U = B\left(f(0), \frac{r}{4}\right) \times B\left(f(1), \frac{r}{4}\right) \times \cdots \times B\left(f(n_0), \frac{r}{4}\right) \times \mathbb{R} \times \mathbb{R}\times \cdots$$

Obviously $f \in U$, and $U$ is a basis element of the product topology. Notice that we also have $U \subseteq B_d(f, r)$. Indeed, for $g \in U$ we have:

$$d(f, g) = \sum_{n=1}^\infty \frac{\min\{1, |f(n) - g(n)|\}}{2^n} < \sum_{n=1}^{n_0}\frac{\min\{1, \overbrace{|f(n) - g(n)|}^{\le \frac{r}4}\}}{2^n} + \frac{r}2 \le r$$

Conversely, let $f \in U_1 \times \cdots \times U_n \times \mathbb{R} \times \mathbb{R} \times \cdots$ be a basis element in the product topology containing $f \in \mathbb{R}^\mathbb{N}$, where $U_1, \ldots, U_n$ are open sets in $\mathbb{R}$.

For $i \in \{1, \ldots, n\}$, since $f(i) \in U_i$, there exists $r_i < 1$ such that $B\left(f, r_i\right) \subseteq U_i$.

Consider $r = \min\left\{r_1, \frac{r_2}{2}, \ldots, \frac{r_n}{2^n}\right\}$, and lets show that $B_d(f, r) \subseteq U_1 \times \cdots \times U_n \times \mathbb{R} \times \mathbb{R} \times \cdots$. Let $g \in B_d(f, r)$. For any $i \in \{1, \ldots, n\}$ we have:

$$\frac{\overbrace{|f(i) - g(i)|}^{<1}}{2^i} = \frac{\min\{1, |f(i) - g(i)|\}}{2^i} \le \sum_{n=1}^\infty \frac{\min\{1, |f(n) - g(n)|\}}{2^n} = d(f, g) < r\le \frac{r_i}{2^i}$$

so $|f(i) - g(i)| \le r_i$ which implies $g(i) \in U_i$.

Now, the product topology on $\mathbb{R}^\mathbb{N}$ has precisely the desired property: a sequence $(f_n)_{n=1}^\infty$ converges in $\mathbb{R}^\mathbb{N}$ if and only if all the coordinate sequences $(f_n(k))_{n=1}^\infty$ converge in $\mathbb{R}$.

Have a look here: Convergence in product topology