Continuity at Every Point
For a function between metric spaces, continuity on its entire domain means continuity at each point of that domain. The delta in the epsilon-delta condition is chosen after the point has been specified, so it may depend on that point as well as on the requested tolerance. This is different from requiring one delta to work simultaneously at every point. That stronger condition is called uniform continuity.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces, each equipped with its metric topology, and let \(f:X\to Y\). The distance from \(f(x)\) to \(f(y)\) is measured by \(d_Y\), while the distance from \(x\) to \(y\) is measured by \(d_X\). The domain need not be all of \(X\) in an ambient space: if a function is defined on a subset, that subset is used as its metric space with the restricted metric.
The difference is in the order of the choices. For continuity on \(X\), first fix a point \(a\); the delta may depend on \(a\). For uniform continuity, the delta must be chosen before the points \(x,y\), and then work for every pair in the domain. The epsilon tolerance is positive in both definitions, and the required conclusion is a strict inequality.
| Condition | What is chosen | Where the estimate must hold |
|---|---|---|
| Continuous at \(a\) | For each \(\varepsilon\), a \(\delta\) that may depend on \(a\) | All \(x\) within \(\delta\) of the fixed point \(a\) |
| Continuous on \(X\) | For each \(\varepsilon\) and each \(a\), a \(\delta\) that may depend on both | All \(x\) within that \(\delta\) of \(a\) |
| Uniformly continuous on \(X\) | For each \(\varepsilon\), one \(\delta\) independent of the points | All pairs \(x,y\in X\) within \(\delta\) of each other |
Global Continuity as a Quantified Condition
Proof. Suppose \(f\) is continuous on \(X\). By definition it is continuous at every \(a\in X\), so for each chosen \(a\) and each \(\varepsilon>0\), the pointwise definition supplies a suitable \(\delta>0\). Conversely, suppose the stated condition holds for every \(a\in X\). Fix any \(a\in X\). The condition gives the epsilon-delta implication required for continuity at this \(a\). Since \(a\) was arbitrary, \(f\) is continuous at every point of \(X\), hence continuous on \(X\). If \(X\) is empty, there are no points at which continuity must be checked, so continuity on \(X\) holds vacuously. \(\square\)
Worked Example: The Square Function on the Real Line
Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), using the standard metric on both copies of \(\mathbb{R}\). We find a delta that proves continuity at an arbitrary fixed point \(a\in\mathbb{R}\).
Fix \(\varepsilon>0\) and \(a\in\mathbb{R}\). The difference factors as \[ |f(x)-f(a)|=|x^2-a^2|=|x-a||x+a|. \] To control the second factor near \(a\), require first that \(|x-a|<1\). Then \[ |x+a|=|(x-a)+2a|\leq |x-a|+2|a|<1+2|a|. \] Choose \[ \delta=\min\left(1,\frac{\varepsilon}{2|a|+1}\right). \] This is positive. If \(|x-a|<\delta\), then \[ |x^2-a^2|=|x-a||x+a|<\delta(2|a|+1)\leq\varepsilon. \] Thus \(f\) is continuous at \(a\). Since the argument applies to every real \(a\), the function is continuous on \(\mathbb{R}\).
Check: At \(a=0\), the formula gives \(\delta=\min(1,\varepsilon)\), and \(|x|<\delta\leq\varepsilon\) implies \(|x^2|=|x|^2<\varepsilon\), since also \(|x|<1\). The choice is allowed to change with \(a\); this calculation has not shown that \(x^2\) is uniformly continuous on \(\mathbb{R}\).
Uniform Continuity Implies Continuity
Proof. Suppose \(f\) is uniformly continuous, and fix any \(a\in X\) and \(\varepsilon>0\). Uniform continuity provides a \(\delta>0\) such that whenever \(x,y\in X\) satisfy \(d_X(x,y)<\delta\), then \(d_Y(f(x),f(y))<\varepsilon\). Apply this implication with \(y=a\). For every \(x\in X\) with \(d_X(x,a)<\delta\), it gives \(d_Y(f(x),f(a))<\varepsilon\). This is continuity at \(a\). Since \(a\) was arbitrary, \(f\) is continuous on \(X\). \(\square\)
The converse need not hold. To see what can go wrong, compare the square function on the whole real line with its restriction to a bounded interval.
Worked Example: The Square Function on a Bounded Interval
Let \(f:[0,1]\to\mathbb{R}\) be \(f(x)=x^2\), with the metrics inherited from the standard metric on \(\mathbb{R}\). For \(x,y\in[0,1]\), we have \(0\leq x+y\leq2\), so \[ |x^2-y^2|=|x-y||x+y|\leq2|x-y|. \] Given any \(\varepsilon>0\), choose \(\delta=\varepsilon/2\). If \(x,y\in[0,1]\) and \(|x-y|<\delta\), then \[ |x^2-y^2|\leq2|x-y|<2\delta=\varepsilon. \] The same delta works for every pair \(x,y\in[0,1]\), so the function is uniformly continuous on this domain.
Check: The estimate used the domain restriction \(x,y\in[0,1]\) to get \(|x+y|\leq2\). For example, at \(x=2\), \(y=3\) on the whole real line, \(|x+y|=5\), so that bound cannot be used there. This check identifies precisely why the interval argument does not establish uniform continuity on all of \(\mathbb{R}\).
Worked Example: The Square Function Is Not Uniformly Continuous on the Real Line
Again take \(f(x)=x^2\), now with domain \(\mathbb{R}\). To disprove uniform continuity, it is enough to find one tolerance for which every positive proposed delta fails. Use \(\varepsilon=1\). Let \(\delta>0\) be arbitrary. Choose \(x>1/\delta\) and set \(y=x+\delta/2\). Then \[ |x-y|=\delta/2<\delta, \] but \[ |f(y)-f(x)|=|(x+\delta/2)^2-x^2| =x\delta+\frac{\delta^2}{4}>1. \] The last strict inequality follows from \(x>1/\delta\), which gives \(x\delta>1\), and from \(\delta^2/4>0\). Thus every proposed delta admits a pair closer than delta whose function values differ by more than 1. No single delta works for this tolerance, so \(f\) is not uniformly continuous on \(\mathbb{R}\).
Check: The points are in the domain because both \(x\) and \(x+\delta/2\) are real. The input distance is strictly less than the proposed delta, while the output distance is strictly greater than the fixed tolerance \(1\), exactly as needed to refute uniform continuity.
A Pointwise-Continuous Function That Is Not Uniformly Continuous
Worked Example: The Reciprocal Function on the Positive Real Line
Let \(f:(0,\infty)\to\mathbb{R}\) be \(f(x)=1/x\). The domain has the metric restricted from the standard metric on \(\mathbb{R}\), and the codomain has the standard metric. First fix any \(a>0\) and any \(\varepsilon>0\). For \(x>0\), \[ \left|\frac1x-\frac1a\right|=\frac{|x-a|}{ax}. \] If \(|x-a|<a/2\), then \(x>a/2\), so \(ax>a^2/2\). Consequently, \[ \left|\frac1x-\frac1a\right|<\frac{2|x-a|}{a^2}. \] Choose \[ \delta=\min\left(\frac a2,\frac{\varepsilon a^2}{2}\right)>0. \] Whenever \(x\in(0,\infty)\) and \(|x-a|<\delta\), the preceding bound gives \[ \left|\frac1x-\frac1a\right|<\frac{2\delta}{a^2}\leq\varepsilon. \] Therefore the reciprocal function is continuous at every \(a>0\), and hence continuous on its domain.
It is not uniformly continuous on \((0,\infty)\). Fix \(\varepsilon=1\) and let \(\delta>0\) be any proposed uniform choice. Set \[ t=\min(\delta/2,1/4),\qquad x=t,\qquad y=2t. \] Both \(x\) and \(y\) are positive. Also \(t<\delta\): if \(t=\delta/2\), this follows from \(\delta>0\); if \(t=1/4\), then \(\delta/2\geq1/4\), so \(\delta\geq1/2>1/4=t\). Hence \(|x-y|=t<\delta\). But \(t\leq1/4\), and direct substitution gives \[ \left|\frac1x-\frac1y\right| =\left|\frac1t-\frac1{2t}\right| =\frac1{2t}\geq2>1. \] Thus this pair violates the required output bound for every proposed delta. The reciprocal function is continuous at every point of its domain but is not uniformly continuous there.
Common Pitfall: Moving the Point-Dependent Delta Outside the Quantifiers
A common error is to calculate a delta near a fixed point and then treat that formula as though it were one number that works on the whole domain. For the square function, the choice \(\delta=\min(1,\varepsilon/(2|a|+1))\) depends on the fixed point \(a\). It proves continuity at each \(a\), but as \(|a|\) grows, this choice can shrink. It does not supply a positive lower bound that works for all real \(a\). The explicit failure above confirms that no such uniform delta exists on \(\mathbb{R}\).
There is also a frequent quantifier mix-up: uniform continuity does not mean that for each pair \(x,y\) one may choose a suitable delta after seeing that pair. Its delta must depend only on \(\varepsilon\), not on either point. Conversely, continuity on \(X\) does not require the same delta at different points. Keeping the order of choices visible prevents both errors.
The distinction matters when a function is used over a whole domain rather than near a single fixed input. Pointwise continuity controls the output near each chosen input; uniform continuity provides one input-distance tolerance that controls the output everywhere at once. The domain is part of the question: the square function has the required uniform control on \([0,1]\), but not on \(\mathbb{R}\), and the reciprocal function loses uniform control as inputs approach \(0\), which is not in its domain.
Check Your Understanding
For each question, keep track of which choices may depend on the point and which must work across the whole domain.
- Write the quantifiers in the definition of continuity at one point \(a\), and then write those for uniform continuity.
- For \(f(x)=x^2\) on \(\mathbb{R}\), use the pointwise estimate in the tutorial to give a delta at \(a=2\) for a specified \(\varepsilon>0\).
- Why does the estimate \(|x^2-y^2|\leq2|x-y|\) prove uniform continuity on \([0,1]\), and what domain condition makes the bound on \(|x+y|\) valid?
- For the reciprocal function, identify the fixed tolerance and the pair of inputs used to disprove uniform continuity.
- Does uniform continuity imply continuity on the domain? Does continuity at every point imply uniform continuity? Justify each answer using the results and examples above.