[proofplan] We argue by contradiction from the negation of [uniform continuity](/pages/1240). This negation gives two [sequences](/pages/1149) in $[a,b]$ whose mutual distance tends to $0$ while their function values remain separated by a fixed positive amount. Compactness of the closed interval, through the [Bolzano-Weierstrass Theorem](/theorems/171), gives a convergent subsequence of one sequence, and the distance estimate forces the paired subsequence to have the same limit. [Continuity](/pages/1199) of $f$ at that common limit then contradicts the fixed separation of the function values. [/proofplan] [step:Negate uniform continuity to construct close points with separated images] Assume, for contradiction, that $f$ is not uniformly continuous on $[a,b]$. By the negation of [uniform continuity](/page/Uniform%20Continuity), there exists $\epsilon_0 > 0$ such that for every $\delta > 0$, there exist $x,y \in [a,b]$ satisfying \begin{align*} |x-y| < \delta \quad \text{and} \quad |f(x)-f(y)| \geq \epsilon_0. \end{align*} For each $n \in \mathbb{N}$, apply this statement with $\delta = 1/n$. This defines sequences \begin{align*} x_\bullet: \mathbb{N} &\to [a,b], & n &\mapsto x_n, \\ y_\bullet: \mathbb{N} &\to [a,b], & n &\mapsto y_n, \end{align*} such that for every $n \in \mathbb{N}$, \begin{align*} |x_n-y_n| < \frac{1}{n} \quad \text{and} \quad |f(x_n)-f(y_n)| \geq \epsilon_0. \end{align*} [guided] We prove this step by spelling out the logical negation of uniform continuity. Uniform continuity of $f$ on $[a,b]$ means that for every $\epsilon > 0$ there exists $\delta > 0$ such that whenever $x,y \in [a,b]$ and $|x-y| < \delta$, one has $|f(x)-f(y)| < \epsilon$. Therefore, if $f$ is not uniformly continuous, there is one positive tolerance $\epsilon_0 > 0$ that cannot be achieved uniformly: for every $\delta > 0$, there exist $x,y \in [a,b]$ with \begin{align*} |x-y| < \delta \quad \text{and} \quad |f(x)-f(y)| \geq \epsilon_0. \end{align*} Now choose a specific sequence of tolerances tending to $0$, namely $\delta = 1/n$ for $n \in \mathbb{N}$. For each $n$, the failed uniform-continuity condition supplies points $x_n,y_n \in [a,b]$. Thus we obtain sequences \begin{align*} x_\bullet: \mathbb{N} &\to [a,b], & n &\mapsto x_n, \\ y_\bullet: \mathbb{N} &\to [a,b], & n &\mapsto y_n, \end{align*} with \begin{align*} |x_n-y_n| < \frac{1}{n} \quad \text{and} \quad |f(x_n)-f(y_n)| \geq \epsilon_0 \end{align*} for every $n \in \mathbb{N}$. The [first inequality](/theorems/2897) says the paired points are forced together; the second says their function values remain separated by at least the fixed amount $\epsilon_0$. [/guided] [/step] [step:Extract a subsequence whose paired points converge to the same limit] Since $x_n \in [a,b]$ for every $n \in \mathbb{N}$, the sequence $(x_n)$ is bounded in $\mathbb{R}$. By the [Bolzano-Weierstrass Theorem](/theorems/628), there exist a strictly increasing map \begin{align*} n_\bullet: \mathbb{N} &\to \mathbb{N}, & j &\mapsto n_j, \end{align*} and a point $c \in \mathbb{R}$ such that $x_{n_j} \to c$. Since each $x_{n_j}$ lies in the closed interval $[a,b]$, the limit satisfies $c \in [a,b]$. For every $j \in \mathbb{N}$, the triangle inequality gives \begin{align*} |y_{n_j}-c| \leq |y_{n_j}-x_{n_j}| + |x_{n_j}-c| < \frac{1}{n_j} + |x_{n_j}-c|. \end{align*} Because $n_j \to \infty$ and $x_{n_j} \to c$, the right-hand side tends to $0$. Hence $y_{n_j} \to c$. [guided] The sequence $(x_n)$ lies in $[a,b]$, so it is bounded in $\mathbb{R}$ because $a \leq x_n \leq b$ for all $n \in \mathbb{N}$. The Bolzano-Weierstrass Theorem applies to bounded real sequences; therefore there is a subsequence indexed by a strictly increasing map \begin{align*} n_\bullet: \mathbb{N} &\to \mathbb{N}, & j &\mapsto n_j, \end{align*} and there is a real number $c \in \mathbb{R}$ such that $x_{n_j} \to c$. Since every term $x_{n_j}$ belongs to $[a,b]$ and $[a,b]$ is [closed](/pages/1145) in $\mathbb{R}$, the limit point belongs to the same interval: $c \in [a,b]$. We now show that the paired subsequence $(y_{n_j})$ converges to the same point. For each $j \in \mathbb{N}$, the triangle inequality in $\mathbb{R}$ gives \begin{align*} |y_{n_j}-c| \leq |y_{n_j}-x_{n_j}| + |x_{n_j}-c|. \end{align*} The construction of the sequences gives $|y_{n_j}-x_{n_j}| < 1/n_j$, so \begin{align*} |y_{n_j}-c| < \frac{1}{n_j} + |x_{n_j}-c|. \end{align*} Because the indices $n_j$ are strictly increasing, $n_j \to \infty$, and hence $1/n_j \to 0$. Also $x_{n_j} \to c$ by construction. Therefore the upper bound tends to $0$, and the squeeze principle for nonnegative real sequences yields $|y_{n_j}-c| \to 0$. This is exactly $y_{n_j} \to c$. [/guided] [/step] [step:Use continuity at the common limit to contradict the fixed separation] The point $c$ belongs to $[a,b]$, and $f$ is continuous on $[a,b]$ by hypothesis. Therefore $f$ is continuous at $c$. Since $x_{n_j} \to c$ and $y_{n_j} \to c$ with all terms in the domain $[a,b]$, the [sequential characterisation of continuity](/theorems/179) gives \begin{align*} f(x_{n_j}) \to f(c) \quad \text{and} \quad f(y_{n_j}) \to f(c). \end{align*} By continuity of subtraction and absolute value on $\mathbb{R}$, \begin{align*} |f(x_{n_j})-f(y_{n_j})| \to |f(c)-f(c)| = 0. \end{align*} This contradicts the construction, which gives \begin{align*} |f(x_{n_j})-f(y_{n_j})| \geq \epsilon_0 \end{align*} for every $j \in \mathbb{N}$, with $\epsilon_0 > 0$. The contradiction shows that the assumption that $f$ is not uniformly continuous is false. Hence $f$ is uniformly continuous on $[a,b]$. [/step]