# compact set always contains its supremum and infimum

Let $K$ be a compact subset of $\mathbb R$. Prove that $\sup K$ and $\inf K$ exist and are in $K$.

My approach: As $K$ is compact, it is bounded. So $\sup K$ and $\inf K$ exists. The reason is that:

Since $K$ is compact, there exist $k_1, \cdots , k_n \in \mathbb R$ such that

$$K \subset \bigcup_{j=1}^n (-k_j,k_j)$$

If $N = \max\{k_1,\cdots, k_n\}$, then $K$ is a subset of $(-N,N)$. Hence $K$ is bounded.
since $K$ is bounded by $-N$ and $N$, $\sup K$ and $\inf K$ exists.

Is this good enough? Is boundedness guaranteed the existence of supremum and infimum?

#### Solutions Collecting From Web of "compact set always contains its supremum and infimum"

Generally, it’s easier to work with open sets than with closed sets. Open sets have less structure: all their points are interior. Closed sets may have two kinds of points: interior and boundary.

So, I would rather look at $\mathbb R\setminus K$. If $\sup K\in \mathbb R\setminus K$, then by openness, there is an interval $(a,b)$ contained in $\mathbb R\setminus K$ and containing $\sup K$. Show that $a$ is an upper bound for $K$, and you have a contradiction.

Is closed condition necessary for $\inf$ and $\sup$ [to be contained in the set]?

It’s essential for the proof (i.e., we can’t just drop it), but it’s not necessary in the sense that some sets contain their $\inf$ and $\sup$ without being closed. For example, $[-2,1) \cup (1,2]$.

A compact set in a metric set is bounded: cover the set by balls of fixed radius $r$, then find a finite subcover by balls of radius $r$. Then consider d: $K \times K \rightarrow \mathbb R$ takes a max $M$ and a min $m$ since $d$ is a continuous function on a compact set . Now consider $d (0,K) : {0} \times K \rightarrow \mathbb R$ , another continuous, Real-valued function on a compact set, which takes a max M’, and a min m’. Then the ball centered at $0^n$ with radius 2(M+M’) contains $K$.

Now, for the limit points. For every connected component of $K$ that has more than a singleton, both $SupS, InfS$ are limit points of $S$. If they were not, and there is $r$ with $B(x,r) \cap K =\$ {} , then $K$ is not connected. In $R^n$ , compact implies closed and bounded (tho the argument in the above paragraphfor boundedness of compact in metric space still holds.) Since , in this case SupS, LimS are limit points, and a closed set contains all its limit points, a compact set contains its sup and Inf. In the case $K$ is not connected, this still holds.

As $K$ is compact, we have that $K$ is bounded. So $\sup K$ and $\inf K$ exists. By definition $\sup K$, for every $n \in N$ exists $x_n \in K$ such that $\sup K- x_n<1/n$ then $\sup K = \lim x_n$ with $x_n \in K$, as K is closed follows that $\sup K \in K$. To inf is analogous.
Ps: Compact ⇒ closed and bounded.