In mathematical analysis, particularly when dealing with functions on metric spaces or subsets of real numbers, the concept of uniform continuity is crucial. A key result connects uniform continuity with boundedness when the domain is a bounded set. This article explores why a uniformly continuous function defined on a bounded set must always be bounded.
Let’s consider a function $f$ that is uniformly continuous on a set $A$. We want to demonstrate that if $A$ is a bounded subset of $mathbb{R}$, then $f$ must be bounded on $A$. We will use proof by contradiction.
Assume, for the sake of contradiction, that $f$ is unbounded on $A$, even though it is uniformly continuous and $A$ is bounded. Without loss of generality, let’s assume that $f$ is unbounded above, meaning $sup_{xin A} f(x) = infty$. (The case where $f$ is unbounded below, $inf_{xin A} f(x) = -infty$, follows a similar argument.)
If $sup_{xin A} f(x) = infty$, this implies that we can find a sequence ${x_n}$ in $A$ such that $f(x_n) to infty$ as $n to infty$. From this sequence, we can extract a subsequence ${y_n}$ of ${x_n}$ such that $f(y_{n+1}) – f(y_n) > 1$ for all $n in mathbb{N}$. This construction ensures that the function values increase by at least 1 along this subsequence.
Since $f$ is uniformly continuous on $A$, for any $epsilon > 0$, there exists a $delta > 0$ such that for all $x, y in A$, if $|x – y| < delta$, then $|f(x) – f(y)| < epsilon$. Let’s choose $epsilon = 1$. Then there exists a $delta > 0$ such that for all $x, y in A$:
$$ |x – y| < delta quad Longrightarrow quad |f(x) – f(y)| < 1. $$
Now, we use the fact that $A subset mathbb{R}$ is bounded. Because $A$ is bounded, any sequence in $A$, including our subsequence ${y_n}$, is also bounded. According to the Bolzano-Weierstrass theorem, the bounded sequence ${y_n}$ must have a convergent subsequence ${z_n}$ that converges to some limit $z$. Since $A$ is a subset of $mathbb{R}$, $z$ will belong to the closure of $A$, denoted as $overline{A}$.
Because ${z_n}$ is a convergent subsequence, it is a Cauchy sequence. This means that for any $delta > 0$, there exists an $N in mathbb{N}$ such that for all $m, n > N$, we have $|z_m – z_n| < delta$. We can choose indices in the subsequence ${z_n}$ such that for any pair of indices $m, n in mathbb{N}$ with $m neq n$, we have $|z_m – z_n| < delta$.
From the uniform continuity condition, this implies that for all $m, n in mathbb{N}$ with $m neq n$, we must have $|f(z_m) – f(z_n)| < 1$.
However, recall that ${z_n}$ is a subsequence of ${y_n}$, which was chosen such that for any consecutive terms in ${y_n}$ (and thus in ${z_n}$ as well, although not necessarily consecutive in ${z_n}$), the difference in function values is greater than 1. This leads to a contradiction because we have $|z_m – z_n| < delta$ while simultaneously requiring $|f(z_m) – f(z_n)| > 1$ (since ${z_n}$ is a subsequence of ${y_n}$ and inherits the property that function values increase by more than 1 along the sequence, though not necessarily for all pairs $z_m, z_n$, but the increasing nature is still inherent and we can select the subsequence ${z_n}$ more carefully to maintain this difference property – or more directly, the contradiction arises from the fact that as $f(y_n) to infty$, for sufficiently large $m,n$, $|f(z_m) – f(z_n)|$ can be made arbitrarily large, contradicting $|f(z_m) – f(z_n)| < 1$).
To refine the contradiction more explicitly, we can reconsider picking the subsequence ${z_n}$ from ${y_n}$ such that $|z_m – zn| < delta$ for all $m, n > N$ for some large N (Cauchy property). Since $f(y{n+1}) – f(y_n) > 1$, for $m$ sufficiently larger than $n$, we will have $|f(z_m) – f(z_n)|$ being a sum of differences each greater than 1, thus $|f(z_m) – f(z_n)| > 1$ for sufficiently separated indices $m, n$. This directly contradicts the uniform continuity condition $|f(z_m) – f(z_n)| < 1$ when $|z_m – z_n| < delta$.
Therefore, our initial assumption that $f$ is unbounded must be false. Hence, a uniformly continuous function on a bounded set $A subset mathbb{R}$ must be bounded.
Extension to the Closure and Compactness
A significant implication of uniform continuity is that a uniformly continuous function $f$ on $A$ can be extended to a continuous function on the closure $overline{A}$. If ${x_n} subset A$ is a Cauchy sequence, then due to the uniform continuity of $f$, ${f(x_n)}$ is also a Cauchy sequence in $mathbb{R}$ and thus converges. This allows for a continuous extension of $f$ to $overline{A}$.
Since $A$ is bounded, its closure $overline{A}$ is also bounded and closed in $mathbb{R}$. A closed and bounded subset of $mathbb{R}$ is compact (Heine-Borel theorem). A continuous function on a compact set is always bounded (Extreme Value Theorem). Therefore, the continuous extension of $f$ to $overline{A}$ is bounded on $overline{A}$, which implies that $f$ itself must be bounded on $A subseteq overline{A}$. This provides another perspective on why uniformly continuous functions on bounded sets are bounded, linking it to the properties of compact sets and continuous functions.
