Continuity Is a Local Condition
The previous tutorial related continuity at an accumulation point to the limit of the function at that point. That viewpoint separates nearby behavior from the value assigned at the point. Here we focus on what continuity at one point forces nearby function values to do. The conclusions are local: they concern inputs in the domain sufficiently close to the specified point, not the function everywhere.
Let \(E\subseteq\mathbb{R}\), let \(f:E\to\mathbb{R}\), and let \(a\in E\). Recall that continuity at \(a\) means that for every \(\varepsilon>0\), there is a \(\delta>0\) such that \(x\in E\) and \(|x-a|<\delta\) imply \(|f(x)-f(a)|<\varepsilon\). The restriction \(x\in E\) matters: when the domain omits points near \(a\), only the inputs it actually contains are relevant.
One useful way to read this definition is to choose a target tolerance and ask what it tells us about all sufficiently nearby inputs. Choosing a tolerance of \(1\), for instance, forces nearby function values to lie within \(1\) of \(f(a)\). Choosing a tolerance smaller than a positive distance to a threshold can force those values to stay on the same side of that threshold.
Continuity Gives Local Boundedness
Proof. Apply the definition of continuity at \(a\) with \(\varepsilon=1\). There is a \(\delta>0\) such that for every \(x\in E\) with \(|x-a|<\delta\), \[ |f(x)-f(a)|<1. \] The triangle inequality gives \[ |f(x)|\leq |f(a)|+|f(x)-f(a)|<|f(a)|+1. \] Thus every value of \(f\) on \(E\cap(a-\delta,a+\delta)\) has absolute value at most \(|f(a)|+1\). This proves boundedness on that relative neighborhood of \(a\). \(\square\)
The theorem says that continuity at just one point is enough to prevent the function from becoming arbitrarily large arbitrarily close to that point. It does not say the function is bounded on its entire domain, or even on every neighborhood of \(a\). The conclusion guarantees one suitable neighborhood.
Worked Example: A Continuous Value at Zero Forces Local Boundedness
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=\sqrt{1+x^2}\). At \(a=0\), we have \(f(0)=1\). To verify continuity there directly, rationalize the difference: \[ |f(x)-f(0)| =\left|\sqrt{1+x^2}-1\right| =\frac{x^2}{\sqrt{1+x^2}+1}. \] The denominator is at least \(2\), so \[ |f(x)-1|\leq \frac{x^2}{2}. \] Given \(\varepsilon>0\), choose \(\delta=\sqrt{2\varepsilon}\). If \(|x|<\delta\), then \(|f(x)-1|\leq x^2/2<\varepsilon\), proving continuity at zero. The local boundedness theorem, or the same estimate with \(\varepsilon=1\), shows that if \(|x|<\sqrt{2}\), then \(|f(x)-1|<1\), and hence \(|f(x)|<2\). Continuity does not require this function to be bounded on all of \(\mathbb{R}\); it gives a bound near zero.
Strict Inequalities Persist Nearby
Continuity also preserves strict inequalities locally. If \(f(a)\) is positive, then sufficiently nearby values remain positive. More generally, if \(f(a)\) lies strictly below or above a number \(c\), then nearby values remain on that same side of \(c\). The size of the allowable neighborhood can depend on the distance between \(f(a)\) and \(c\).
Proof. First suppose \(f(a)<c\), and set \(\varepsilon=c-f(a)\), which is positive. By continuity, there is a \(\delta>0\) such that \(x\in E\) and \(|x-a|<\delta\) imply \(|f(x)-f(a)|<c-f(a)\). In particular, \[ f(x)-f(a)\leq |f(x)-f(a)|<c-f(a), \] so \(f(x)<c\). Now suppose \(f(a)>c\), and set \(\varepsilon=f(a)-c>0\). Continuity gives a \(\delta>0\) such that \(|f(x)-f(a)|<f(a)-c\) for the specified nearby inputs. Therefore \[ f(a)-f(x)\leq |f(x)-f(a)|<f(a)-c, \] which implies \(f(x)>c\). Both claims follow. \(\square\)
Taking \(c=0\) shows that a continuous function that is positive at \(a\) remains positive on some relative neighborhood of \(a\). If \(f(a)\ne0\), the theorem likewise shows that the function cannot change sign sufficiently close to \(a\). This is a pointwise conclusion: the neighborhood may be small, and it is chosen for the particular point and threshold.
Worked Example: Keeping Nearby Values Above a Threshold
Let \(f(x)=5-2x\) on \(\mathbb{R}\), and consider \(a=1\) and the threshold \(c=2\). Since \(f(1)=3>2\), the theorem guarantees that \(f(x)>2\) for all \(x\) sufficiently close to \(1\). We can verify a specific choice directly: \[ f(x)-2=3-2x=1-2(x-1). \] If \(|x-1|<1/4\), then \(2|x-1|<1/2\), and consequently \[ f(x)-2=1-2(x-1)\geq 1-2|x-1|>\frac12>0. \] Thus \(|x-1|<1/4\) ensures \(f(x)>2\). The strict gap \(f(1)-2=1\) is what allows a neighborhood in which the inequality persists.
Only Nearby Agreement Matters
For continuity at \(a\), values far from \(a\) have no effect. This can be stated precisely for two functions on the same domain. If the functions agree at every domain point sufficiently close to \(a\), including \(a\) itself, then either both are continuous at \(a\) or neither is.
Proof. Suppose \(f\) is continuous at \(a\), and let \(\varepsilon>0\). There is a \(\delta_1>0\) such that \(x\in E\) and \(|x-a|<\delta_1\) imply \(|f(x)-f(a)|<\varepsilon\). Set \(\delta=\min\{\delta_1,r\}\). If \(x\in E\) and \(|x-a|<\delta\), the assumed local agreement gives both \(g(x)=f(x)\) and \(g(a)=f(a)\). Hence \[ |g(x)-g(a)|=|f(x)-f(a)|<\varepsilon. \] This proves that \(g\) is continuous at \(a\). The same argument with \(f\) and \(g\) exchanged proves the reverse implication. \(\square\)
Worked Example: Changing a Function Far from the Point
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=x^2\), and define \(g:\mathbb{R}\to\mathbb{R}\) by \[ g(x)= \begin{cases} x^2, & |x|<1,\\ 7, & |x|\geq 1. \end{cases} \] The functions agree whenever \(|x|<1\), so the local agreement theorem says that they have the same continuity behavior at \(0\). To check it directly, \(g(0)=0\), and for \(|x|<1\) we have \(g(x)=x^2\). Given \(\varepsilon>0\), choose \(\delta=\min\{1,\sqrt{\varepsilon}\}\). If \(|x|<\delta\), then \(|g(x)-g(0)|=x^2<\varepsilon\). Thus \(g\) is continuous at zero, even though its values were changed at inputs outside the interval \((-1,1)\). Those changes cannot affect continuity at zero.
Pointwise Continuity Does Not Control the Whole Neighborhood
Local boundedness and persistence of strict inequalities are consequences of continuity at a point, but continuity at that point does not mean the function is continuous at every nearby point. A function may have well-behaved values as its inputs approach \(a\) while behaving irregularly at other points in every neighborhood of \(a\).
Worked Example: Continuity at Zero Without Continuity Everywhere Nearby
Define \(h:\mathbb{R}\to\mathbb{R}\) by \[ h(x)= \begin{cases} x, & x\in\mathbb{Q},\\ 2x, & x\notin\mathbb{Q}. \end{cases} \] In particular, \(h(0)=0\). For every real \(x\), either \(h(x)=x\) or \(h(x)=2x\), so \(|h(x)|\leq2|x|\). Given \(\varepsilon>0\), choose \(\delta=\varepsilon/2\). Then \(|x|<\delta\) implies \(|h(x)-h(0)|=|h(x)|\leq2|x|<\varepsilon\). Thus \(h\) is continuous at zero.
However, \(h\) is not continuous at any nonzero point \(a\). To see why, choose rational numbers \(r_n\) and irrational numbers \(s_n\) such that both sequences converge to \(a\); such choices are possible because both the rationals and the irrationals are dense in \(\mathbb{R}\). Then \(h(r_n)=r_n\to a\), whereas \(h(s_n)=2s_n\to2a\). Since \(a\ne0\), these limits differ. The sequential test for limits from the previous tutorial therefore rules out continuity at \(a\). Every neighborhood of zero contains nonzero points, so \(h\) is continuous at zero but is not continuous throughout any neighborhood of zero.
How to Use Pointwise Continuity
When applying continuity at a point, first identify the precise local conclusion needed. If the goal is to bound function values, use the local boundedness theorem. If the goal is to ensure a function remains positive or stays below a threshold, use persistence of strict inequalities. If a formula has been altered away from the point, check whether it agrees with the original formula on some relative neighborhood; if so, continuity at the point is unchanged.
These tools also provide quick tests for failure. If a function is unbounded in every neighborhood of \(a\), it cannot be continuous there, because continuity would imply local boundedness. Likewise, if \(f(a)>c\) but values \(f(x)\leq c\) occur arbitrarily close to \(a\), the strict-inequality theorem shows that \(f\) cannot be continuous at \(a\). Such arguments can be simpler than trying to find a suitable \(\delta\) for every \(\varepsilon\).
All these statements are relative to the domain. A neighborhood in the conclusions means the points \(x\in E\) close to \(a\), not every real number in an interval if some of those numbers are outside \(E\). At an isolated point, continuity is automatic, as discussed in the previous tutorial; the local conclusions are then compatible with the fact that there may be no other domain inputs to check nearby.
Check Your Understanding
Use the local consequences of continuity to answer the following questions.
- Which value of \(\varepsilon\) gives a direct proof that a function continuous at \(a\) is bounded on some relative neighborhood of \(a\)?
- If \(f\) is continuous at \(a\) and \(f(a)>c\), why must \(f(x)>c\) for all domain inputs sufficiently close to \(a\)?
- If two functions agree on \(E\cap(a-r,a+r)\) for some \(r>0\), what can be concluded about their continuity at \(a\)?
- Why does local boundedness provide a test that can disprove continuity at a point?
- Can a function be continuous at \(a\) but discontinuous at other points in every neighborhood of \(a\)? Explain using the example in this tutorial.