Prove that $[a,b]$ is compact

Question

Let $a<b$ be real numbers. Prove that $[a,b]$ is compact.

Below I present my solution. I thought it's good enough, but my TA said it's incorrect. I don't see where there is a problem. Could you help me find it?

My proof:

Let $\{U_i\}_{i\in I}$ be an open cover of $[a,b]$. Let $$S=\{x\in [a,b]: x>a \mbox{ and } [a,x] \mbox{ is covered by a finite union of sets from }\{U_i\} \} \mbox{.}$$

It is clear that $S$ is bounded. Moreover, $S$ is not empty because there is $j\in I$ such that $a\in U_j$. Since $U_j$ is open, there is some $\epsilon>0$ such that $[a,\epsilon) \subset U_j$, therefore $a+\frac{\epsilon}{2}\in S$.

We have showed that $S$ is bounded and not empty so there exists $\sup S$. We claim that $\sup S=b\in S$.

It's obvious that $\sup S \in U_k$ for some $k\in I$ and $\sup S \le b$. Assume $c=\sup S<b$. Then there is $\delta>0$ such that $[c, c + \delta)\subset U_k$ and $[c, c + \delta)\subset [a,b]$. But this means that we can cover $[a,c + \frac{\delta}{2}]$ by finitely many sets from $\{U_i\}$, so $c$ is not the supremum. Thus it must be the case that $\sup S=b$.

Answer 1

You have showed that $S$ is a subset of $[a,b]$ with minimum $a$ and supremum $b.$ By definition of $S,$ one can (and should) prove that either $S=[a,b)$ or $S=[a,b].$ You must rule out $S=[a,b).$

Answer 2

Let $ G_\alpha $ be an open cover of $[a,b]$ with no finite subcover. Then divide this interval in exactly two halves and assume that one of them is not covered by finitely many $G_\alpha$'s since otherwise we will have a finite subcover. Now call this interval $A_1$ and $A_0=[a,b]$. Now divide $A_1$ in exactly two halves and by same argument we can the interval $A_3$ with no finite subcover . continuing in this way we get a sequence of nested intervals $A_0 , A_1 , A_2$, and so on. We see that if $x , y \in A_n$, then $|x-y| \le \frac{|a-b|}{2^n}$.

By the nested interval property there is a point in the intersection of $A_n$'s, then $m \in \mathbb{N}$ for which $A_m$ is contained in some $G_\alpha$ (why?) And this is a contradiction

Answer 3

A subset of $\mathbb{R}$ is compact iff it is closed and bounded. If we set $$M=\max \{|a|,|b|\},$$ then $\forall\, x \in[a, b]$ it automatically follows $|x| \leq M$, with which we have already shown that the set $[a, b]$ is bounded.

Now let the interval $[a, b]$ be closed. We have to prove $x \in[a, b]$, that is $a \leq x \leq b$. Since $x_n \in[a, b]$, that is $a \leq x_n \leq b$, holds for every $n \in \mathbb{N}$, by the sandwich criterion for sequences at the limit $n \rightarrow+\infty$ it just follows $a \leq \lim _n x_n \leq b$ and therefore $x \in[a, b]$ as desired. Thus the interval $[a, b]$ is compact.

(Sandwich Criterion) Let $I$ be an interval containing a value $a$. Let $f, g$ and $h$ be functions defined on $I \backslash\{a\}$. If for each $x \neq a$ from $I$ holds $$ g(x) \leq f(x) \leq h(x), $$ as well as $$ \lim _{x \rightarrow a} g(x)=\lim _{x \rightarrow a} h(x)=L, $$ then $$ \lim _{x \rightarrow a} f(x)=L . $$