Krdytkyu

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 $$U (A,epsilon) := { x : d(x,A) < epsilon }.$$
  

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• (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.

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
$$ U(mathbb{N}, epsilon) = bigcup_{nin mathbb{N}} (n-epsilon, n+epsilon). tag{2}$$

And so we have
$$ U(mathbb{N}, epsilon) notsubseteqq U. tag{3} $$
To see why , let us choose a natural number $N > frac{1}{2epsilon}$. Then $epsilon > frac{1}{2N}$, and so the open interval $left( N-epsilon, N+epsilon right)$ is not contained in the open interval $left( N - frac{1}{2N}, N+frac{1}{2N} right)$; moreover the latter open interval is the only one of the disjoint open intervals in set $U$ [Please refer to (Definition 1) above.] which has a point in common with the former open interval. Since the open interval $left( N-epsilon, N+epsilon right)$ is contained in $U(mathbb{N}, epsilon)$, 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
$$ U ( mathbb{N}, epsilon ) = bigcup_{n in mathbb{N} } big( n-epsilon, n+epsilon big) supset bigcup_{n in mathbb{N} } big( n-delta, n+delta big) = U( mathbb{N}, delta ). $$
Note that $0 < delta < frac{1}{2}$.
And, by (3) as
$$ U(mathbb{N}, delta) notsubseteqq U, $$
so we can also conclude that $$ U( mathbb{N}, epsilon ) notsubseteqq U $$
either.

Is this example a suitable one?

Your example is fine. Another simple one, which I think allows to visualize what happens, is as follows: let $A={(x,0)| xin mathbb{R}}$ 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 $xrightarrow infty$, so that no neighbourhood of $A$ with diameter bounded from below will fit into it.

Yes, your example is nice. (If $0notinmathbb N$.)