This question concerns exercise 2(e) from section 27 (p.177) in Munkres' Topology:
Let $(X,d)$ be a metric space, and let $A$ be a non-empty subset of $X$.
- For any point $x \in X$, we define $$d(x, A) := \inf \{ \ d(x,a) \ \colon \ a \in A \ \}.$$
- For $\epsilon > 0$ we define the $\epsilon$-neighbourhood of $A$ in $X$ to be the set $$U (A,\epsilon) := \{ \, x \in X \, : \, d(x,A) < \epsilon \, \}.$$
- (d) Assume that $A$ is compact; let $U$ be an open set containing $A$. Show that some $\epsilon$-neighborhood of $A$ is contained in $U$.
- (e) Show the result in (d) need not hold if $A$ is closed but not compact.
Here is a Math Stack Exchange post of mine on Part (d).
My effort:
Consider the set $\mathbb{N}$ in $\mathbb{R}$ with the usual metric. Then $\mathbb{N}$ is closed, but not bounded and hence not compact.
Let the set $U$ be given by
$$U \colon= \bigcup_{n \in \mathbb{N}} \left( n – \frac{1}{2n}, n + \frac{1}{2n} \right). \tag{Definition 1 } $$This set $U$ is an open set containing $\mathbb{N}$. We also note that set $U$ is a union of disjoint open intervals.
Now let $\epsilon$ be any positive real number.
Let us first choose $\epsilon$ such that
$$ 0 < \epsilon < \frac{1}{2} . $$
Then by Part (c) of this problem, we can show that
$$ U(\mathbb{N}, \epsilon) = \bigcup_{n\in \mathbb{N}} (n-\epsilon, n+\epsilon). \tag{2}$$And so we have
$$ U(\mathbb{N}, \epsilon) \not\subseteqq U. \tag{3} $$
To see why , let us choose a natural number $N$ such that
$$ N > \frac{1}{2\epsilon}. $$
Then
$$ \epsilon > \frac{1}{2N}, $$
and if $r$ is any real number such that
$$ \epsilon > r > \frac{1}{2N}, $$
then we find that the real number $N + r$, for example, satisfies
$$ N+r \in \{ N- \epsilon, N + \epsilon \} \setminus \left( N – \frac{1}{2N}, N + \frac{1}{2N} \right), $$
and so the open interval $\left( N-\epsilon, N+\epsilon \right)$ is not contained in the open interval $\left( N – \frac{1}{2N}, N+\frac{1}{2N} \right)$, that is,
$$ ( N- \epsilon, N + \epsilon ) \not\subset \left( N – \frac{1}{2N}, N + \frac{1}{2N} \right); $$
moreover as the latter open interval $\left( N – \frac{1}{2N}, N + \frac{1}{2N} \right)$ is the only one of the disjoint open intervals in set $U$ that has a point in common with the former open interval $( N- \epsilon, N + \epsilon )$ [Please refer to (Definition 1) above.], so we can conclude that
$$ (N – \epsilon, N + \epsilon) \not\subset U, $$
and since the open interval $\left( N-\epsilon, N+\epsilon \right)$ is contained in $U(\mathbb{N}, \epsilon)$ by virtue of (2) above, we can conclude that (3) does indeed hold.Now for any real $\epsilon > 0$, we can a choose a real number $\delta$ such that
$$ 0 < \delta < \min \left\{ \ \epsilon, \ \frac{1}{2} \ \right\}. $$
Then
$$ U ( \mathbb{N}, \epsilon ) = \bigcup_{n \in \mathbb{N} } \big( n-\epsilon, n+\epsilon \big) \supset \bigcup_{n \in \mathbb{N} } \big( n-\delta, n+\delta \big) = U( \mathbb{N}, \delta ). $$
That is,
$$U ( \mathbb{N}, \epsilon ) \supset U( \mathbb{N}, \delta ). \tag{4} $$
Note that $0 < \delta < \frac{1}{2}$.
And, by (3) as
$$ U(\mathbb{N}, \delta) \not\subseteqq U, $$
so we can also conclude that $$ U( \mathbb{N}, \epsilon ) \not\subseteqq U $$
either.
Is this example a suitable one?
Best Answer
Your example is fine. Another simple one, which I think allows to visualize what happens, is as follows: let $A=\{(x,0)| x\in \mathbb{R}\}$ the real line embedded in $\mathbb{R}^2$. Then let $$U := \{(x,y)| |y|< \frac{1}{|x|}\}$$ (with the convention $1/0 := \infty$) the region between the graphs of $\pm 1/x$. The graph of $1/x$ approaches $A$ as $x\rightarrow \infty$, so that no neighbourhood of $A$ with diameter bounded from below will fit into it.