Closed Real Interval is Compact

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Theorem

Let $\R$ be the real number line considered as an Euclidean space.

Let $I = \closedint a b$ be a closed real interval.


Then $I$ is compact.


Topological Space (Proof 1)

This proves that $I$ is compact in the context of a general topological space:


Let $\UU$ be any open cover of $\closedint a b$.

Let:

$G = \set {x \in \R: x \ge a, \closedint a x \text{ is covered by a finite subset of } \UU}$

Let points in $G$ be classified as $\text{good}$ (for $\UU$).

Thus if $z$ is $\text{good}$, then $\closedint a z$ has that finite subcover that is to be demonstrated for the whole of $\closedint a b$.

The aim therefore is to show that $b$ is $\text{good}$.


Let $x$ be $\text{good}$, and $a \le y \le x$.

We have that $\closedint a y \subseteq \closedint a x$.

Thus $\closedint a y$ can be covered with any finite subset of $\UU$ that covers $\closedint a x$.

As $x$ be $\text{good}$, at least one such subset is known to exist.

Thus, by definition, $y$ is $\text{good}$.


Now we show that $G \ne \O$.

At the same time we show that $G \supseteq \hointr a {a + \delta}$ for some $\delta > 0$.


As $\UU$ covers $\closedint a x$, it follows that $a$ must belong to some $U \in \UU$.

So, let $U \in \UU$ be some open set such that $a \in U$.

Since $U$ is open, $\hointr a {a + \delta} \subseteq U$ for some $\delta > 0$.

Hence $\closedint a x \subseteq U$ for all $x \in \hointr a {a + \delta}$.

It follows that all these $x$ are $\text{good}$.

So $G \ne \O$, and:

$(1): \quad G \supseteq \hointr a {a + \delta}$ for some $\delta > 0$


Now the non-empty set $G$ is either bounded above or it is not.


Suppose $G$ is not bounded above.

Then there is some $c$ which is $\text{good}$ such that $c > b$.

From our initial observation that if $x$ is $\text{good}$, and $a \le y \le x$, then $y$ is $\text{good}$, it follows that $b$ is $\text{good}$, and hence the result.


Now suppose $G$ is bounded above.

By the Continuum Property, $G$ admits a supremum in $\R$.

So let $g = \sup G$.


Suppose that $g > b$.

As $g$ is the least upper bound for $G$, there exists some $c$ which is $\text{good}$ such that $c > b$.

Again, from our initial observation, it follows that $b$ is $\text{good}$, and hence the result.


Aiming for a contradiction, suppose that $g \le b$.

From $(1)$ above:

$\hointr a {a + \delta} \subseteq G$ for some $\delta > 0$

and so $g > a$.

Since $g \in \closedint a b$, $g$ must belong to some $U_0 \in \UU$.

Since $U_0$ is open, there exists some open $\epsilon$-ball $\map {B_\epsilon} g$ of $g$ such that $U_0 \supseteq \map {B_\epsilon} g$.

Since $g > a$, we can arrange that $\epsilon < g - a$.

As $g$ is the least upper bound, there must be a $\text{good}$ $c$ such that $c > g - \epsilon$.

This means $\closedint a c$ is covered by a finite subset of $\UU$, say $\set {U_1, U_2, \ldots, U_r}$.

Then $\closedint a {g + \dfrac \epsilon 2}$ is covered by $\set {U_1, U_2, \ldots, U_r, U_0}$.

So $g + \dfrac \epsilon 2$ is $\text{good}$, contradicting the fact that $g$ is an upper bound for $G$.

This contradiction implies that $g > b$, and hence $b \in G$.


Since $\UU$ was arbitrary, we conclude that every open cover of $\closedint a b$ has a finite subcover.

This satisfies the topological definition of compact, thus the proof is complete.

$\blacksquare$



This page has been identified as a candidate for refactoring.
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Topological Space (Proof 2)

This proves that $I$ is compact in the context of a general topological space:

The space $A_n = \struct {\set {0, 1}, \tau_n}$ given the discrete topology is compact.

By Tychonoff's theorem, $\ds A = \prod_{n \mathop = 1}^\infty A_n$ is compact, where $A$ has the product topology.

By Cantor Space as Countably Infinite Product, $A$ is homeomorphic to the Cantor space, $\CC$

Thus Cantor space is compact. By the definition as a ternary representation, $\CC$ consists of all the elements of $\closedint 0 1$ which has an expansion in base $3$ without using $1$.

Let $x \in \CC$. Then in base $3$ notation, we have (as $0 \le x \le 1$):

$\ds x = \sum_{j \mathop = 1}^\infty a_j 3^{-j}$

where $\forall j: a_j \in \set {0,2}$.

From, Representation of Ternary Expansions, the expansion is unique.

Define the following function:

$\ds f: \CC \to \closedint 0 1, \quad \map f {\sum_{j \mathop = 1}^\infty a_j 3^{-j} } = \sum_{j \mathop = 1}^\infty \frac {a_j} 2 2^{-j}$

Note that $\forall j: \dfrac{a_j} 2 \in \set {0,1}$.

Let $j \in \N$. Note that if $x,y \in \CC$ such that $|x-y| < 3^{-j}$, then $|f(x)-f(y)| \le 2^{-j}$.

Thus $f$ is continuous.

From Existence of Base-N Representation, every element of $y \in \closedint 0 1$ can be written as

$\ds y = \sum_{j \mathop = 1}^\infty b_j 2^{-j}$

where $\forall j: b_j \in \set {0,1}$.

For $x \in \CC$ given by

$\ds x = \sum_{j \mathop = 1}^\infty 2 b_j 3^{-j}$

$f(x)=y$

So $f: \CC \to \closedint 0 1$ is a continuous surjection.

By Continuous Image of Compact Space is Compact, $\closedint 0 1$.

Since $I$ is homeomorphic to $\closedint 0 1$, $I$ is compact.

$\blacksquare$

Metric Space

This proves that $I$ is compact in the context of a metric space:


From Closed Real Interval is Closed Set, $I$ is a closed set of $\R$.

From Real Interval is Bounded in Real Numbers, $I$ is bounded in $\R$.

The result follows by definition of compact.

$\blacksquare$


Normed Vector Space

This proves that $I$ is compact in the context of a normed vector space:


We have that $\struct {\R, \size {\, \cdot \,}}$ is a normed vector space.

Let $\sequence {a_n}_{n \mathop \in \N}$ be a sequence in $I$.

$\sequence {a_n}$ is bounded by $a$ and $b$.

By Bolzano-Weierstrass theorem, there exists a convergent subsequence $\sequence {a_{n_k}}_{k \in \N}$ with a limit $L$.

Then:

$\forall k \in \N : a \le a_{n_k} \le b$

Take the limit $k \to \infty$:

$a \le L \le b$

Hence, $L \in I$.

The result follows by definition of compact.

$\blacksquare$


Also see


Sources

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