Heine borel theorem doesn't hold

Question

Let $X$ be a metric space with underlying set $\mathbb{R}$ and metric $d(x,y)=\min\{|x-y|,1\}$. We have to find a subset of $X$ which is closed and bounded but not compact.

My attempt:

Since $d(x,y)=\min\{|x-y|,1\}$ , it means every set is bounded. Now choose a subset $X_n=\lbrace1/n, n\in N\rbrace$ is this closed? To prove it is not compact take an open cover of $X_n=(1/n,1/2)$ now it doesn't have finite subcover, why?

Answer 1

Simply take the whole $\mathbb R$. It is closed and bounded, but not compact. For instance, the sequence $1,2,3,\ldots$ has no convergent subsequence. Here, I am using the fact that a metric space is compact if and only if every sequence of elements of that space has a convergent subsequence.

Answer 2

Why not just choose the whole space? This is closed, and it is bounded because the metric is, and it clearly isn't compact because the open cover by open balls of the form $(\frac n4-0.5, \frac n4+0.5) $ has no finite subcover.

Answer 3

Critique of the Attempt:

I find the provided attempt to be impenetrable—I do not understand what the asker is trying to say. At first, it appears that the asker is claiming that the singleton set $X_n = \{1/n\}$ is closed and bounded, but not compact. Yes, this set is closed and bounded, but it is also compact: if $\mathscr{U} = \{U_\alpha\}_{\alpha \in A}$ is an open cover of $\{1/n\}$ (where $A$ is some (possibly infinite) index set), then for any fixed $a\in A$, the collection $\{ U_a \}$ (i.e. the singleton collection) is an open subcover of $\{1/n\}$.

In the next sentence, the asker seems to redefine $X_n$ to be the interval $(1/n,1/2)$. Presumably, the asker is seeking to cover some closed and bounded set by these intervals. However, the argument doesn't seem to go anywhere, and it is not clear what the asker intends.

In short, the attempt does not clearly define a closed and bounded set $C$, nor does it clearly define an open cover $\mathscr{U}$ of that set $C$. In order to cogently argue that $(X,d)$ possesses a subset which is closed and bounded but not compact by way of example, it is necessary to first define a set with the desired properties, then demonstrate that these properties hold.

Demonstration:

As others have noted, the entire space is closed and bounded, but not compact. Let us now demonstrate that $\mathbb{R}$ has the claimed properties:

• By definition of a topology, a space is always both open and closed in itself. Therefore $\mathbb{R}$ is closed with respect to the topology generated by the metric.

• The space is bounded because $$ \mathbb{R} = B(0,2) = \{ x\in \mathbb{R} : d(x,0) < 2 \}. $$ In slightly more detail, if $x \in \mathbb{R}$, then $$ d(x,0) = \min\{|x-0|,1\} = \min\{ |x|, 1 \} \le 1 < 2. $$ Therefore $x\in B(0,2)$. Indeed, if $r > 1$, then the ball of radius $r$ centered at any point in $\mathbb{R}$ will contain all of $\mathbb{R}$.

  Because $\mathbb{R}$ is contained in a ball of finite radius it is, by definition, bounded. Note that this further implies that every subset of $\mathbb{R}$ is also bounded with respect to this topology.

• The space $\mathbb{R}$ is not compact. For example, define the open cover $$ \mathscr{U} := \left\{ B(\tfrac{n}{2}, \tfrac{1}{2}) \right\}_{n\in\mathbb{Z}}, $$ i.e. the set of balls of radius $\tfrac{1}{2}$ centered at points of the form $\tfrac{n}{2}$ where $n$ is an integer. Note that we might equivalently write $\mathscr{U}$ as the collection of intervals $$ \mathscr{U} = \left\{ (\tfrac{n}{2}-\tfrac{1}{2}, \tfrac{n}{2}+\tfrac{1}{2}) \right\}_{n\in\mathbb{Z}}. $$ However, I think that it is more enlightening to think of the intervals as open balls: the metric topology is generated by open balls, which immediately implies that the balls themselves are open. On the other hand, it is possible to define a metric (or, more generally, a topology) on $\mathbb{R}$ such that "open" intervals are not open.

  I think that it is clear that $\mathscr{U}$ covers $\mathbb{R}$. While a more rigorous argument can be provided, I think that it is sufficient to note that the subcollection $$ \left\{ B(n, \tfrac{1}{2}) \right\}_{n\in\mathbb{Z}} $$ covers all of $\mathbb{R}$ except for points of the form $\tfrac{2n+1}{2}$ where $n$ is an integer (that is, the only points not covered by this subcollection odd multiples of $\tfrac{1}{2}$). However, the odd multiples of $\tfrac{1}{2}$ comprise the centers of the remaining balls which make up $\mathscr{U}$, hence $\mathscr{U}$ covers $\mathbb{R}$.

  Finally, note that $\mathscr{U}$ has no finite subcover. Indeed, if we remove any set from $\mathscr{U}$, it will no longer cover $\mathbb{R}$, as the center of the removed ball will not be contained in any of the remaining sets.

Comments:

In the demonstration above, I chose to cover the real line by intervals of length $1$ centered at multiples of $\tfrac{1}{2}$. The reason to chose such a cover is pedagogical: it should not take very much work to rigorously convince someone that the intervals are open, it is reasonably clear that the intervals cover $\mathbb{R}$, and if any interval is removed from this cover then the new collection no longer covers $\mathbb{R}$. The other answers are, perhaps, more elegant, but they have the potential to run afoul of technicalities which might be confusing or surprising to a student who is just being introduced to these topological concepts.

For example, prior to editing, Matt Samuel suggested covering $\mathbb{R}$ by sets of the form $(-n,n)$. This, indeed, is sufficient to show that $\mathbb{R}$ is not compact. However, there is a lurking technicality which should be considered: the set $(-n,n)$ is not an open ball with respect to the funny metric which has been defined for this space. It is open, as it can be written as the union of open balls, e.g. \begin{align} (-n,n) &= \bigcup_{k=-n+1}^{n-1} B(k, 1) \\ &= \bigcup_{k=-n+1}^{n-1} (k-1,k+1) \\ &= (-n,-n+2) \cup (-n+2, -n+4) \cup \dotsb \cup (n-4,n-2) \cup (n-2, n). \end{align} Once we are happy that $(-n,n)$ is open, it is not too hard to show that the collection $\{(-n,n)\}_{n\in\mathbb{Z}}$ has no finite subcover.

José Carlos Santos' solution is also nice, but it relies on a theorem which relates compactness (defined by the property that every cover has a finite subcover) with sequential compactness (every sequence has a convergent subsequence). This theorem is not too difficult to prove, but some proof is necessary.

Answer 4

In this metric every set is bounded. So just find a set that is closed but would be unbounded under the metric $d'(x,y) = |x-y|$. (BTW, Is there an assumption that $X \subset \mathbb R$?; defining $d(x,y)=\min(|x-y|, 1)$ doesn't make sense if $x-y$ and $|x-y|$ aren't defined for $x,y \in X$)

So for instance that $X= \mathbb R$ and the $\mathbb R$ itelf is bounded. And closed as it is the entires space. Let $U = \{(x,x+1)|x\in \mathbb R$ but the open cover. That has no finite subcover.