Basic Analysis: Limits and Continuity (3)

Let’s consider multivariate functions f : D\to \mathbf{R}^m where D\subseteq \mathbf{R}^n. To that end, we need the Euclidean distance function on Rn. If x = (x1x2, …, xn) is in Rn, we define:

|\mathbf{x}| := \sqrt{x_1^2 + x_2^2 + \ldots + x_n^2}.

Note that |x| = 0 if and only if x is the zero vector 0. Now we are ready to define continuity.

[ In this and all subsequent articles, boldfaced variables axy, … denote parameters with multiple components. ]

Definition. The function f:D\to\mathbf{R}^m is said to be continuous at a if:

  • for every ε>0, there exists δ>0, such that whenever |xa|<δ and x belongs to D, we have |f(x)-f(a)|<ε.

Let D\subseteq \mathbf{R}^n. A point \mathbf{a}\in\mathbf{R}^n is said to be a point of accumulation, if for any ε>0, there exists \mathbf{x}\in \mathbf{R}^n such that

0 < |\mathbf{x}-\mathbf{a}|<\epsilon.

Theorem. If a is a point of accumulation of D, and we have continuous functions

g, h : D\cup\{\mathbf{a}\}\to \mathbf{R}^m

which agree on D-{a}, then g(a)=h(a).

We’ll skip the proof since it’s identical to the case of single-variable functions, but we’d encourage the conscientious reader to check that the logic follows through.

Basic Properties

Continuous functions satisfy the following basic properties:

Theorem. The property of continuity satisfies the following.

  1. Identity. The identity map D\to D\subseteq\mathbf{R}^n is continuous.
  2. Restriction. If D'\subseteq D and f:D\to\mathbf{R}^m is continuous at a in D’, then so is the restriction of f to D’.
  3. Composition. If f:D\to E is continuous at a and g:E\to\mathbf{R}^m is continuous at b=f(a), then g\circ f:D\to \mathbf{R}^m is continuous at a.
  4. Projection. The map \pi_i : \mathbf{R}^m \to\mathbf{R} which projects to i-th component is continuous at every point.
  5. Reduction of Codomain to R. The function f:D\to\mathbf{R}^m is continuous at a iff \pi_i\circ f is continuous for each i.
  6. Arithmetic. The addition and product functions \mathbf{R}\times\mathbf{R}\to\mathbf{R} are continuous, as is the reciprocal \mathbf{R}-\{0\}\to\mathbf{R}.


All the properties have rather straightforward proofs. We’ll use our earlier lingo: say “when x close to a in D, P(x) holds” if there exists δ>0 such that whenever |xa|<δ and lies in D, the property P(x) holds.

  1. Identity. Easy: let δ=ε.
  2. Restriction. Again let δ=ε.
  3. Composition. Let ε>0. Since g is continuous at f(a), there exists δ>0 such that when |yf(a)|<δ and y is in E|g(y)-gf(a)|<ε. Likewise, since f is continuous at a, there exists δ’>0 such that when |xa|<δ’ and x is in D, then |f(x)-f(a)|<δ and hence |gf(x)-gf(a)|<ε.
  4. Projection. Use the inequality |\pi_i(\mathbf{x})-\pi_i(\mathbf{a})| \le |\mathbf{x}-\mathbf{a}|.
  5. Reduction of Codomain to R. If f is continuous, then \pi_i\circ f is continuous by the above two properties. Conversely, suppose each \pi_i\circ f is continuous. Let ε>0. For x close to a in D, we have |\pi_i f(\mathbf{x})-\pi_i f(\mathbf{a})|< \epsilon/\sqrt{m} for each i=1,2,…,m. Hence, when x is close to a in D, we have |f(x)-f(a)| < ε as well.
  6. Arithmetic.
    • For addition, use the inequality |(x+y)-(a+b)| ≤ |xa|+|yb| ≤ |(x,y)-(a,b)|.
    • For multiplication, |xyab| = |x(yb)+(xa)b| ≤ |x|·|yb|+|b|·|xa|. When |xa|<1, we then have |x|<|a|+1. Hence, for any ε>0, whenever |xa|<min(1, ε/(2|a|+2)) and |yb|<ε/(2|b|) we have |xyab|<ε. Hence set δ=min(1, ε/(2|a|+2), ε/(2|b|)).
    • For division, take |\frac 1 x-\frac 1 a| = \frac{|a-x|}{|a|\cdot|x|}. When |xa|<|a|/2, we have |x|>|a|/2. Hence, for any ε>0, whenever |xa|<min(|a|/2, ε|a|2/2) we get |\frac 1 x-\frac 1 a|<\epsilon.


Prove the following using only the above properties, without using the ε-δ definition.

  1. Concatenation” of continuous functions gives a continuous function. Specifically, suppose f:D\to \mathbf{R} and f:E\to\mathbf{R} are continuous at and a’ respectively, where D\subseteq \mathbf{R} and E\subseteq\mathbf{R}. Then h=(f,g):D\times E\to\mathbf{R}^2 given by \mathbf{a}\mapsto (f(\mathbf{a}), g(\mathbf{a})) is continuous at (a, a’).
  2. Prove that if f, gD → R are continuous at a, for D\subseteq \mathbf{R}, then f+gf×g are also continuous at a. If f(a)≠0, then 1/f is continuous at a also. Hence, any polynomial function is continuous.

Answers (highlight to read).

  1. Consider D×→ D → R, where the first map is projection (and hence continuous at (aa’) ) and the second map is f. Since it’s a composition of two continuous maps, the result is continuous. Likewise, D×→ E → is continuous. On the other hand, the two maps are also given by composing (fg): D×→ R×R → R, where the 2nd map is a projection. The 4th property then tells us (fg) is continuous at (aa’).
  2. Use property 5: the composition D×D → R×R → R, where the first map is the concatenation (fg) and the second map is addition/product, is continuous.

Global Continuity

We say that a function f:D\to \mathbf{R}^m is continuous if it’s continuous at every point in the domain. From definition, it’s clear that continuity at a specific point is determined by what happens close to that point. To that end, let’s define open sets.

Definitions. Let D\subseteq \mathbf{R}^n be any subset. An open ball containing \mathbf{a}\in D is the set of points:

N(\mathbf{a}, \epsilon)=\{\mathbf{x}\in D : |\mathbf{x}-\mathbf{a}|<\epsilon\},

for some ε>0. [ Note that the definition depends on D, although the notation doesn’t reflect that. If there’s a possibility of confusion, we’ll write it as N_D(\mathbf{a},\epsilon). ]

A subset U\subseteq D is said to be an open subset of D if for each \mathbf{a}\in D, we have N(\mathbf{a}, \epsilon)\subseteq U for some ε>0.

E.g. each open ball U := N(a, ε) is open, for if x is in U, then |xa|<ε, so we let δ = ε-|xa|. We claim that N(\mathbf{x},\delta)\subseteq N(\mathbf{a},\epsilon). Indeed, any y in N(x, δ) satisfies |yx|<δ so

|ya| ≤ |yx| + |xa| < δ+|xa| = ε.

More Examples of Open Subsets

  1. Let DR. Then any open interval (bc) is an open subset of D since it’s an open ball. The punctured neighbourhood (bc) – {a} is also an open subset of D since it’s a union of two open balls. The half-open interval U = (bc] is not open in D since the point c in U isn’t contained in any open ball of R which is a subset of U. [ Any open ball containing c is of the form (c-ε, c+ε), which contains points outside U. ]
  2. Let D = [0, 1]\cup\{2\}. The subset U = [0, 1] is now an open subset of D. This may confuse the reader at first glance, but note that even if we pick the endpoint 1 (in U), we can pick the open ball N(1, 1/2) in D. By definition, this is the set \{x\in D: |x-1|<\frac 1 2\}, so it’s (1/2, 1] which is wholly contained in U. Likewise, the singleton subset V = {2} is also a subset of D, since the open ball N(2, 1/2) = {2} is wholly contained in U.
    • Which of the subsets [0, 1), (0, 1], {1}, [0, 1/2), (0, 1/2] are open in D? [ Answer: the first, second, fourth. ]

Here are further examples of open subsets U\subseteq D, where D is coloured in blue and U in red.

In the first two examples, D=\{(x,y)\in\mathbf{R}^2: x^2+y^2\le 1\} while in the other two, D=\{(x,y)\in\mathbf{R}^2:x^2 + y^2\le 1\}\cup\{(2,0)\}. The dotted boundary lines indicate that the boundaries are not included in U.

The main theorem we wish to prove is:

Theorem. Let f:D\to \mathbf{R}^m be a function, where D\subseteq \mathbf{R}^n. Then f is continuous if and only if for each open subset U\subseteq \mathbf{R}^m, f^{-1}(U) is an open subset of D.


Suppose f is continuous. To show that f^{-1}(U) is open in D, let \mathbf{a}\in f^{-1}(U). Since f(a) is in U which is open, there exists an ε>0 such that N(f(a), ε) is a subset of U. And since f is continuous at a, there exists a δ>0 such that whenever |xa|<δ and x in D, we have |f(x)-f(a)|<ε. Hence

f(N(\mathbf{a}), \delta))\subseteq N(f(\mathbf{a}, \epsilon)\subseteq U\implies N(\mathbf{a},\delta)\subseteq f^{-1}(U)

which shows that f^{-1}(U) is open in D.

Conversely, suppose f^{-1}(U)\subseteq D is open for any open subset U\subseteq \mathbf{R}^m. To show f is continuous at a (an element of D), let ε>0. Since U:=N(f(\mathbf{a}),\epsilon)\subset \mathbf{R}^n is open, so is f^{-1}(U). But \mathbf{a}\in f^{-1}(U), so by definition of open subsets, there exists δ>0 for which

N(\mathbf{a}, \delta)\subseteq f^{-1}(U) \implies f(N(\mathbf{a},\delta))\subseteq U=N(f(\mathbf{a}),\epsilon)

so whenever |xa|<δ and x in D, we have |f(x)-f(a)|<ε. ♦

Main point. The above theorem suggests that the central concept behind analysis is that of open subsets. Indeed, the core idea behind topology is to build up an entire theory based only on the concept of open subsets. This will be elaborated in subsequent posts.

Important Exercises.

  1. Find a continuous function fR → R and an open subset U of R such that f(U) is not open in R.
  2. Prove that if U and V are open subsets of D, then so is U ∩ V. In particular, this means that finite intersections of open subsets are still open.
  3. Prove that if {Ui} is a collection of open subsets of D, then so is U:=\cup_i U_i.
  4. Suppose E\subseteq D.
    1. Prove that if U is open in D, then V := ∩ E is open in E.
    2. Prove that conversely, if V is open in E, then V∩ E for some open subset U of D.

[ Note: for Q4, be careful with the notation N(a, ε) since the ambient space can be either D or E. ]

Sketch of Answers [Highlight to read.]

  1. Let f(x)=x2 and U=(-1, 1). Then f(U) = [0, 1) is not open in R. However, in this case, it is open in the image of f, which is [0, ∞). Can you find an example where f(U) is not open in the image of f?
  2. Let a be in U ∩ V. For some ε>0 and ε’>0, N(a, ε) and N(a, ε’) are subsets of U and V respectively. Thus, N(a, min(ε, ε’)) is a subset of U ∩ V.
  3. Let a be in U. Then a lies in some Ui so for some ε>0, N(a, ε) is a subset of Ui and hence U.
  4. For (A), suppose a is in V. So a is in the open set U and for some ε>0, ND(a, ε) is a subset of U. Since NE(a, ε) = ND(a, ε) ∩ E, we also have NE(a, ε) is a subset U ∩ EV. For (B), let a be in V, which is open in E. For some ε>0, NE(a, ε) is a subset of V. Now let U be the union of all open balls ND(a, ε) in D. Each open ball is open in D and by Q3, U is also open in D. Clearly VU ∩ E.
This entry was posted in Notes and tagged , , , , , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s