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?


1 Answer 1


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.


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