Here is Prob. 2 (b), Sec. 25, in the book Topology by James R. Munkres, 2nd edition:
Consider $\mathbb{R}^\omega$ in the uniform topology. Show that $\mathbf{x}$ and $\mathbf{y}$ lie in the same component of $\mathbb{R}^\omega$ if and only if the sequence $$ \mathbf{x} - \mathbf{y} = \left( x_1 - y_1, x_2 - y_2, \ldots \right) $$ is bounded. [Hint: It suffices to consider the case where $\mathbf{y} = \mathbf{0}$.]
My Attempt:
First of all, we note that $\mathbb{R}^\omega$ denotes the set of all the (infinite) sequences of real numbers; that is, $\mathbb{R}^\omega$ denotes the countably infinite Cartesian product $$ \mathbb{R} \times \mathbb{R} \times \cdots. $$
And, the uniform topology on $\mathbb{R}^\omega$ is the one induced by the so-called uniform metric on $\mathbb{R}^\omega$, which is given by $$ \bar{\rho}( \mathbf{x}, \mathbf{y} ) \colon= \sup \left\{ \ \min \left\{ \ \left\lvert x_n - y_n \right\rvert, 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \right\} $$ for all $\mathbf{x} \colon= \left( x_n \right)_{n \in \mathbb{N}}$ and $\mathbf{y} \colon= \left( y_n \right)_{n \in \mathbb{N}}$ in $\mathbb{R}^\omega$.
Now let $\mathbf{a} \colon= \left( a_n \right)_{n \in \mathbb{N}}$ and $\mathbf{b} \colon= \left( b_n \right)_{n \in \mathbb{N}}$ be any two points in $\mathbb{R}^\omega$ such that the sequence $$ \mathbf{a} - \mathbf{b} = \left( a_n - b_n \right)_{n \in \mathbb{N}}$$ is bounded, that is, such that there exists a positive real number $r$ such that $$ \left\lvert a_n - b_n \right\rvert \leq r \ \mbox{ for all } \ n \in \mathbb{N}. \tag{1} $$ Now let $f \colon [ 0, 1 ] \to \mathbb{R}^\omega$ be the map defined by $$ f(t) \colon= \mathbf{a} + t(\mathbf{b}- \mathbf{a} ) = \big( a_n + t \left( b_n - a_n \right) \big)_{n \in \mathbb{N}} \ \mbox{ for all } \ t \in [0, 1]. $$ Then for any points $s, t \in [0, 1]$, we note that $$ \begin{align} \bar{\rho}( f(s), f(t) ) &= \sup \left\{ \ \min \left\{ \ \big\lvert \big( a_n + s \left( b_n - a_n \right) \big) - \big( a_n + t \left( b_n - a_n \right) \big) \big\rvert, \, 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \right\} \\ &= \sup \left\{ \ \min \left\{ \ \lvert s-t \rvert \cdot \left\lvert b_n - a_n \right\rvert, \, 1 \ \right\} \ \colon \ n \in \mathbb{N} \ \right\} \\ &\leq \sup \left\{ \ \lvert s-t \rvert \cdot \left\lvert b_n - a_n \right\rvert \colon \ n \in \mathbb{N} \ \right\} \\ &= \lvert s-t \rvert \cdot \sup \left\{ \ \left\lvert b_n - a_n \right\rvert \colon \ n \in \mathbb{N} \ \right\} \\ &\leq \lvert s-t \rvert r \qquad \mbox{ [ using (1) above ]}. \end{align} $$ So, given any real number $\varepsilon > 0$, if we choose our real number $\delta$ such that $$ 0 < \delta < \frac{\varepsilon}{r}, $$ then we find that $$ \bar{\rho}( f(s), f(t) ) < \varepsilon $$ holds for any pair of points $s, t \in [0, 1]$ for which $$ \lvert s-t \rvert < \delta. $$ Thus the map $f$ is continuous (in fact uniformly continuous). Moreover, $f(0) = \mathbf{a}$ and $f(1) = \mathbf{b}$. Therefore $f$ is a path in the uniform space $\mathbb{R}^\omega$ joining the points $\mathbf{a}$ and $\mathbf{b}$. So $\mathbf{a}$ and $\mathbf{b}$ lie in the same path component of $\mathbb{R}^\omega$. And since each path component of any topological space is contained in a component of that space, we can conclude that $\mathbf{a}$ and $\mathbf{b}$ lie in the same component of $\mathbb{R}^\omega$.
Am I right? Can we give an independent proof of the above without having recourse to path components?
Now suppose that the sequence $\mathbf{a} - \mathbf{b}$ is unbounded. Then, for any natural number $k$, there exists a natural number $n_k$ such that $$ \left\lvert a_{n_k} - b_{n_k} \right\rvert > k. $$
Now let $U$ be the set of all the points $\mathbf{x} \colon= \left( x_n \right)_{n \in \mathbb{N}}$ in $\mathbb{R}^\omega$ such that the sequence $\mathbf{a} - \mathbf{x}$ is bounded. Then $\mathbf{a} \in U$.
Moreover, If $\mathbf{u} \colon= \left( u_n \right)_{n \in \mathbb{N}} \in U$, then the sequence $\left( a_n - u_n \right)_{n \in \mathbb{N}}$ is bounded; that is, there exists a positive real number $r_\mathbf{u}$ such that $$ \left\lvert a_n - u_n \right\rvert < r_\mathbf{u} \ \mbox{ for all } \ n \in \mathbb{N}. $$ So if $\varepsilon \in (0, 1)$ and if $\mathbf{x}$ is any point of $\mathbb{R}^\omega$ such that $$ \bar{\rho}( \mathbf{x}, \mathbf{u} ) < \varepsilon, $$ then, for each $n \in \mathbb{N}$, we have $$ \min\big\{ \ \left\lvert x_n - u_n \right\rvert, \ 1 \ \big\} \leq \bar{\rho}( \mathbf{x}, \mathbf{u} ) < \varepsilon < 1, $$ and so $$ \min\big\{ \ \left\lvert x_n - u_n \right\rvert, \ 1 \big\} = \left\lvert x_n - u_n \right\rvert, $$ and thus $$ \left\lvert x_n - u_n \right\rvert < \varepsilon. $$ Then, for any $n \in \mathbb{N}$, we obtain $$ \begin{align} \left\lvert x_n - a_n \right\rvert &\leq \left\lvert x_n - u_n \right\rvert + \left\lvert u_n - a_n \right\rvert \\ &< \varepsilon + r_\mathbf{u}. \end{align} $$ thus showing that $\mathbf{x} \in U$ also. Therefore $U$ is open in the uniform space $\mathbb{R}^\omega$.
Now the point $\mathbf{b}$ is in the set $\mathbb{R}^\omega \setminus U$.
Let $\mathbf{v}$ be any point of $\mathbb{R}^\omega \setminus U$. Then the sequence $\left( v_n - a_n \right)_{n \in \mathbb{N}}$ is unbounded; so for any natural number $k$ there exists a natural number $m_k$ such that $$ \left\lvert v_{m_k} - a_{m_k} \right\rvert > k. $$ Now if $\varepsilon \in (0, 1)$, and if $\mathbf{y}$ is any point of $\mathbb{R}^\omega$ such that $$ \bar{\rho}( \mathbf{y}, \mathbf{v} ) < \varepsilon, $$ then as before we must have $$ \left\lvert y_n - v_n \right\rvert < \varepsilon $$ for every natural number $n$. Now if $\lambda$ is an arbitrary real number and if $k$ is a natural number such that $k > \lambda + \varepsilon$, then we find that $$ \begin{align} \left\lvert y_{m_k} - a_{m_k} \right\rvert &\geq \left\lvert v_{m_k} - a_{m_k} \right\rvert - \left\lvert y_{m_k} - v_{m_k} \right\rvert \\ &> k - \varepsilon \\ &> \lambda. \end{align} $$ Thus corresponding to any real number $\lambda$, no matter how large, we can find a natural number $m_k$ such that $$ \left\lvert y_{m_k} - a_{m_k} \right\rvert > \lambda. $$ Thus the sequence $\left( y_n - a_n \right)_{n \in \mathbb{N}}$ is unbounded, which shows that $\mathbf{y}$ is in $\mathbb{R}^\omega \setminus U$. Therefore the set $\mathbb{R}^\omega \setminus U$ is also open in the uniform metric space $\mathbb{R}^\omega$.
Thus the sets $U$ and $\mathbb{R}^\omega \setminus U$ constitute a separation of the uniform metric space $\mathbb{R}^\omega$. Moreover, these two sets are both open and closed in $\mathbb{R}^\omega$ with the uniform topology.
Finally, if $\mathbf{x}$ and $\mathbf{y}$ are any two points of $U$. then the sequences $\mathbf{x} - \mathbf{a}$ and $\mathbf{y} - \mathbf{a}$ are both bounded; so the sequence $\mathbf{x} - \mathbf{y} = ( \mathbf{x} - \mathbf{a} ) - ( \mathbf{y} - \mathbf{a} )$ is bounded as well. Thus $\mathbf{x}$ and $\mathbf{y}$ are in the same (path) component of the uniform metric space $\mathbb{R}^\omega$, as has been shown above. Thus the set $U$ is (path) connected.
Similarly, the set $V$ of all the points $\mathbf{v}$ in $\mathbb{R}^\omega$ such that $\mathbf{v} - \mathbf{b}$ is bounded is (path) connected. Moreover, $\mathbf{b}$ is in $V$ but $\mathbf{a}$ is not in $V$. Also $V$ is both open and closed in the uniform metric space $\mathbb{R}^\omega$, just as $U$ has been shown to be both open and closed.
Moreover, the sets $U$ and $V$ are disjoint.
I'm getting lost now!
What next? How to proceed from here? Or, is there anywhere I've gone wrong?
P.S.:
I think I've just managed to hit upon the right trick!
Now as the sets $U$ and $\mathbb{R}^\omega \setminus U$ form a separation of the uniform metric space $\mathbb{R}^\omega$, so any connected subspace lies in either $U$ or $\mathbb{R}^\omega \setminus U$, but not both. [Refer to Lemma 23.2 in Munkres.]
As $\mathbf{a} \in U$, so the component containing $\mathbf{a}$, which is of course a connected subspace of $\mathbb{R}^\omega$, must also lie in $U$. And, as $\mathbf{b}$ is in $\mathbb{R}^\omega \setminus U$, so the component containing $\mathbf{b}$ lies in $\mathbb{R}^\omega \setminus U$. Therefore the points $\mathbf{a}$ and $\mathbf{b}$ lie in different components of $\mathbb{R}^\omega$ in the uniform topology if the sequence $\mathbf{a} - \mathbf{b}$ is unbounded.
Is there any flaw in my reasoning?
