1
$\begingroup$

Here is Prob. 4, Sec. 29, in the book Topology by James R. Munkres, 2nd edition:

Show that $[0, 1]^\omega$ is not locally compact in the uniform topology.

Here is a Math Stack Exchange (MSE) post that is of course relevant. However, here I would like to present my own attempt:

First of all, here is an MSE post of mine on $[0, 1]^\omega$ with the uniform topology.

Suppose if possible that $[0, 1]^\omega$ with the uniform metric topology is locally compact.

Let $$ \mathbf{a} \colon= \left( \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \ldots \right). \tag{Definition 0} $$

As $[0, 1]^\omega$ is locally compact at point $\mathbf{a}$, so there exists a compact subspace $C$ of $[0, 1]^\omega$ and an open set $U$ in $[0, 1]^\omega$ such that $$ \mathbf{a} \in U \subset C. \tag{0} $$

Now as $U$ is an open set in the uniform metric space $[0, 1]^\omega$ and as $\mathbf{a} \in U$, so there exists a real number $\delta > 0$ such that $$ B ( \mathbf{a}, \delta ) \subset U, \tag{1} $$ where $$ B ( \mathbf{a}, \delta ) \colon= \{ \, \mathbf{x} \in [0, 1]^\omega \, \colon \, \bar{\rho}( \mathbf{x}, \mathbf{a} ) < \delta \, \}. \tag{ Definition 1 } $$ Since reducing $\delta$ will make the set $B ( \mathbf{a}, \delta )$ smaller, we can assume without any loss of generality that our $\delta$ satisfies $$ 0 < \delta < \frac{1}{2}. \tag{1*} $$

From (0) and (1) above we also obtain $$ B ( \mathbf{a}, \delta ) \subset C. \tag{2} $$

Since $C$ is a compact subspace of the Hausdorff space $[0, 1]^\omega$ with the uniform metric topology, therefore $C$ is also closed in $[0, 1]^\omega$, by Theorem 26.3 in Munkres.

Now as $C$ is a closed set in $[0, 1]^\omega$ and as $B ( \mathbf{a}, \delta ) \subset C$ by (2) above, so we also have $$ \overline{B ( \mathbf{a}, \delta ) } \subset C, $$ that is, $$ \bar{B} ( \mathbf{a}, \delta ) \subset C, \tag{3} $$ where $$ \bar{B} ( \mathbf{a}, \delta ) \colon= \{ \, \mathbf{x} \in [0, 1]^\omega \, \colon \, \bar{\rho}( \mathbf{x}, \mathbf{a} ) \leq \delta \, \}. \tag{ Definition 2} $$

Moreover, as $\bar{B} ( \mathbf{a}, \delta )$ is a closed set in the compact space $C$, so $\bar{B} ( \mathbf{a}, \delta )$ is also compact, by Theorem 26.2 in Munkres.

Finally, as $\bar{B} ( \mathbf{a}, \delta )$ is a compact (metrizable) space, so it is also limit point compact, by Theorem 28.1 in Munkres.

Now let us take $$ \alpha \colon= \frac{1}{2} - \frac{\delta}{2}, \qquad \mbox{ and } \qquad \beta \colon= \frac{1}{2} + \frac{\delta}{2}. \tag{Definition 3*} $$ And, then let us define the set $A$ as $$ A \colon= \{ \, \alpha, \beta \, \}^\omega. \tag{Definition 3 } $$ Then $A$ is an infinite subset of $\bar{B} ( \mathbf{a}, \delta )$, but $A$ has no limit points in $\bar{B} ( \mathbf{a}, \delta )$, as has been shown in my post here. This contradicts the fact that $\bar{B} ( \mathbf{a}, \delta )$ is limit point compact.

Thus our supposition at the start of this proof is wrong. Hence $[0, 1]^\omega$ in the uniform topology is not locally compact.

Is this proof correct? Is it easy enough to understand? Or, are there issues of accuracy or clarity?

$\endgroup$

1 Answer 1

2
$\begingroup$

I think the proof is fine, and easy enough. It's a generalisation of the idea to show the unit ball in $\ell^\infty$ not being compact.

Such infinite-dimensional linear-like spaces will almost never be locally compact.

$\endgroup$

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.