# Prob. 2(b), Sec. 25, in Munkres' TOPOLOGY, 2nd ed: The iff-condition for two points to be in the same component of $\mathbb{R}^\omega$

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?

• Does your notational distinction between $\mathbf R$ and $\mathbb R$ carry any significance? If so, it should be explained; if not, it should be removed. – joriki Jul 28 '18 at 3:00
• @joriki I write $\mathbb{R}$ to denote the set of real numbers. I'll just edit my post to omit $\mathbf{R}$ whereever I've used it. Thanks for the comment. – Saaqib Mahmood Jul 28 '18 at 11:38
• See also math.stackexchange.com/q/2730869. – Paul Frost Jul 28 '18 at 13:51

1. A hint was provided in the exercise, which is that it is sufficient to consider the case where one of the sequence is always vanishing. Indeed it simplifies the notations. You can prove that considering this case is enough by considering the translation $\tau_a$ such that $\tau_a(\mathbf 0)= \mathbf a$. $\tau_a$ is an homeomorphism of $\mathbb R^\omega$, and therefore transform components into components.
2. Regarding your question to avoid using path connectedness. You can notice that the sup norm is well defined on $U$. Also, the open ball $B$ centered on $\mathbf a$ with radius $2\Vert \mathbf a -\mathbf b\Vert_\infty$ is open, connected as all open balls and included in $U$. Therefore $U$ is connected. This is a general result that a union of connected sets that all contain a point $x$ is connected.