2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 '18 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.