Here is Prob. 8, Sec. 23, in the book Topology by James R. Munkres, 2nd edition:

Determine whether or not $\mathbb{R}^\omega$ is connected in the uniform topology.

My Attempt:

By definition, $\mathbb{R}^\omega$ is the set of all the sequences of real numbers.

We define a sequence $\left( x_n \right)_{n \in \mathbb{N} }$ of real numbers to be bounded if there exists a real number $r$ such that $$ \left\lvert x_n \right\rvert < r $$ for all $n \in \mathbb{N}$.

Let $B$ be the set of all the bounded sequences of real numbers, and let $U$ be the set of all the unbounded sequences of real numbers.

Then both $B$ and $U$ are non-empty subsets of $\mathbb{R}^\omega$, and we also have $$ \mathbb{R}^\omega = B \cup U. $$

The uniform metric on $\mathbb{R}^\omega$, by definition, is given by the formula $$ \bar{\rho} \left( \left( x_n \right)_{n \in \mathbb{N} } , \left( y_n \right)_{n \in \mathbb{N} } \right) \colon= \sup \left\{ \ \min \left\{ \left\lvert x_n - y_n \right\rvert, 1 \right\} \ \colon \ n \in \mathbb{N} \ \right\} $$ for any elements $\left( x_n \right)_{n \in \mathbb{N} } , \left( y_n \right)_{n \in \mathbb{N} } $ of $\mathbb{R}^\omega$.

We now show that both $B$ and $U$ are open in the metric space $\left( \mathbb{R}^\omega, \bar{\rho} \right)$.

First, for the set $B$ of all the bounded sequences of real numbers: Let $\mathbf{p} \colon= \left( p_n \right)_{n \in \mathbb{N} } \in B$. Then there is a real number $r$ such that $$ \left\lvert p_n \right\rvert < r \mbox{ for all } n \in \mathbb{N}. $$ Now let us take any real number $\delta$ such that $0 < \delta < 1$. If $\mathbb{x} \colon= \left( x_n \right)_{ n \in \mathbb{N} } \in B ( \mathbf{p}, \delta )$, then $\mathbf{x} \in \mathbb{R}^\omega$ and $$ \bar{\rho} ( \mathbf{x}, \mathbf{p} ) < \delta. $$ Then, for each $n \in \mathbb{N}$, we have $$ \min \left\{ \left\lvert x_n - p_n \right\rvert, 1 \right\} \leq \bar{\rho} ( \mathbf{x}, \mathbf{p} ) < \delta < 1, $$ which implies that $$ \min \left\{ \left\lvert x_n - p_n \right\rvert, 1 \right\} = \left\lvert x_n - p_n \right\rvert. $$ Thus we can conclude that, for each $n \in \mathbb{N}$, we have
$$ \left\lvert x_n - p_n \right\rvert < \delta, \tag{0} $$ and so $$ \left\lvert x_n \right\rvert \leq \left\lvert x_n - p_n \right\rvert + \left\lvert p_n \right\rvert < \delta + r, $$ from which it follows that $\mathbf{x} \in B$ also.

Thus we have shown that, for any point $\mathbf{p} \in B$, we can find a real number $\delta > 0$ such that the open ball $B( \mathbf{p}, \delta )$ in the metric space $\left( \mathbf{R}^\omega, \bar{\rho} \right)$ is contained in $B$. Therefore $B$ is open in $\left( \mathbf{R}^\omega, \bar{\rho} \right)$.

Now for the set $U$ of all the unbounded sequences of real numbers: Let $\mathbf{p} \colon= \left( p_n \right)_{n \in \mathbb{N} }$ be an arbitrary point of $U$. Then, for any real number $r$, we can find a natural number $n_r$ such that $$\left\lvert p_{n_r} \right\rvert > r. \tag{A} $$

Let us again take our $\delta$ such that $0 < \delta < 1$.

Let $r$ be an arbitrary real number. We show that this $r$ is not an upper bound for any sequence $\mathbf{x} \colon= \left( x_n \right)_{n \in \mathbb{N} } \in B( \mathbf{p}, \delta)$.

If $\mathbf{x} \in B( \mathbf{p}, \delta)$, then $\mathbf{x} \in \mathbb{R}^\omega$ and also $\bar{\rho}( \mathbf{x}, \mathbf{p} ) < \delta$, which again implies that, for each $n \in \mathbb{N}$, we have
$$ \left\lvert x_n - p_n \right\rvert < \delta, $$ as in (0) above.

In particular, for $n = n_{r+1}$ in (A) above, we have $$ \left\lvert x_{n_{r+1} } - p_{n_{r+1} } \right\rvert < \delta, $$ and therefore $$ \left\lvert p_{n_{r+1} } \right\rvert \leq \left\lvert p_{n_{r+1} } - x_{n_{r +1} } \right\rvert + \left\lvert x_{n_{r+1} } \right\rvert, $$ which implies that $$ \left\lvert x_{n_{r+1} } \right\rvert \geq \left\lvert p_{n_{r+1} } \right\rvert - \left\lvert p_{n_{r+1} } - x_{ n_{r+1} } \right\rvert > r + 1 - \delta > r, $$ which shows that no real number $r$ can be an upper bound for the sequence $\left( x_n \right)_{n \in \mathbb{N} }$ and hence this sequence is unbounded. Therefore, $\mathbf{x} \in U$ also.

Thus it follows that $U$ is also open in $\left( \mathbb{R}^\omega, \bar{\rho} \right)$.

Is this solution correct? If so, then is the presentation hereof correct as well?


1 Answer 1


As the $\delta$'s are irrelevant, just pick a fixed $\delta=\frac{1}{2}$ and show that $B(\mathbf{p}, \frac{1}{2}) \subseteq B$ for any $\mathbf{p} \in B$ ( I agree with that part) and also that $B(\mathbf{p}, \frac{1}{2}) \subseteq U$ for any $\mathbf{p} \in U$.

In that last proof you picked $r$ before $x$ in the ball, which is not quite correct. Just say, let $\mathbf{x} \in B(\mathbf{p}, \frac{1}{2})$, and pick $r \in \mathbb{R}$, to show unboundedness. There is some $n_r$ such that $|p_{n_r}| \ge r+1$ by unboundedness of $\mathbf{p}$ and then use $|x_{n_r} - p_{n_r}| < \frac{1}{2}$ to see that $|x_{n_r}| \ge r$, as required. But the idea is fine.

  • $\begingroup$ thank you for your answer. I've just edited my post. So can you please have a look at it now and point out if there is still any mistake? $\endgroup$ Mar 15, 2018 at 14:37

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