# ε-δ Definition of Continuous Function

Remember the ε-δ definition of Continuous Function that you have learnt in your high school ?

CT1. A function f : R → R is said to be continuous at x0 ∈ R if
∀ε>0, ∃δ>0 such that|x−x0|<δ ⇒|f(x)−f(x0)|< ε

A function is continuous over an open interval if it is continuous at every point in the interval.

Take the function as shown in the above diagram. When x0 = 2 and ε = 0.5, there is a open interval (1.5, 2.5) for x such that|f(x) -f(2)| < 0.5. So, this function is continuous at 2.

Consider another function, Unit Step Function, which is a discontinuous one:

When x0 = 0 and ε = 0.1, ∀δ > 0, ∀x ∈ (-δ, δ)⇒|f(x) -f(0)| = 0.5 > ε. Thus, the continuous condition is not satisfied and function is discontinuous at x=0.

# Topological Definition of Continuous Function

We can generalise the definition of Continuous Function from Euclidean space R to the more general Topological space. To be more concrete, the condition “|x−x0|<δ ” can be generalised by considering the point x is within the open set defined for point x0. In terms of topology language, we can express the ε-δ definition of Continuous Function as follows:

Let (X, τX) be topological space for set X and (Y, τY) be topological space for set Y. N(p) denotes Neighbourhood of p, please see General Topology-Part 3 for its definition.

CT2. A mapping f : X → Y is said to be Continuous if
∀ oY ∈ τY, f-1(oY) ∈ τX

CT3. A mapping f : X → Y is said to be Continuous at p ∈ X if
∀ Ny ∈ N(f(p)), ∃ Nx ∈ N(p) s.t. f(Nx) ⊆ Ny

A mapping is said to be Continuous if it is Continuous ∀ p ∈ X

In topological terms, “mapping” instead of “function” will be used. Also, there is no distance concept.

Let’s prove that both definitions are equivalent. In this case, τX and τY will be the standard topology on R: τ ={o|o = ∪(a,b)}. (Please refer to my previous article to recap the definition of topological space.)

Proof-
CT1⇒CT2:
∀oY ∈ τY
⇒ oY = ∪(a,b) [ Open set is a union of bases which are (a,b) for R ]
⇒ f-1(oY) = f-1(∪(a,b)) = ∪f-1((a,b))
⇒ f-1(oY) = ∪f-1({y| |y-y0| < ε where ε=(b-a)/2>0 and y0=(a+b)/2})

if f-1(oY) ≠ ø
⇒ f-1(oY) = ∪{ x| |x-x0| < δ ⇒ |f(x)−f(x0)| < ε where f(x0)=y0} [ By ε-δ definition ]
⇒ f-1(oY) = ∪(x0-δ, x0+δ) ∈ τX

if f-1(oY) = ø
⇒ f-1(oY) ∈ τX [By T1, ø is an open set]

CT2⇒CT1:
∀ε>0, |f(x)−f(x0)|< ε, let oY =(f(x0)-ε, f(x0)+ε)
⇒ oY ∈ τY
⇒ ∃oX ∈ τX s.t. x0 ∈ oX and oX= f-1(oY)) [ By Topological definition ]
⇒ ∃oX ∈ τX s.t. x0 ∈ oX and oX= ∪(a[i],b[i])) [ Open set is a union of bases which is (a,b) for R ]
⇒ ∃(a, b) ⊆ ∪(a[i],b[i])) s.t. x0 ∈ (a, b) ⊆ oX
⇒ ∃δ=min(x0-a, b-x0)>0 s.t. x ∈ (x0-δ, x0+δ) ⊆ (a, b)

To test the continuity of a map from a topological space on X to that on Y, sometimes it will be more easy to use basis or subbasis of topological space on Y.

Let’s prove that the following statements are equivalent:
CT2. A mapping f : X → Y is continuous if
∀ oY ∈ τY, f-1(oY) ∈ τX
CT3. A mapping f : X → Y is said to be Continuous at p ∈ X if
∀ Ny ∈ N(f(p)), ∃ Nx ∈ N(p) s.t. f(Nx) ⊆ Ny
CT4. Inverse image of every basis element of τY is open.
CT5. Inverse image of every subbasis element of τY is open.

Proof-
CT2 ⇒ CT3
∀ Ny ∈ N(f(p))
⇒ ∃ oY ∈ τY s.t. f(p) ∈ oY ⊆ Ny [ By CT2 ]
⇒ ∃ oX=f-1(oY) ∈ τX where p ∈ oX
⇒ ∃ oX ∈ N(p) s.t. f(oX)=oY ⊆ Ny

CT3 ⇒ CT2
∀ oY ∈ τY
⇒ ∀y=f(p) ∈ oY, ∃ Nx ∈ N(p) s.t. f(Nx) ⊆ oY [ By CT3, oY∈ N(y) ]
⇒ ∃ oX[p] ∈ τX s.t. p ∈ oX[p] ⊆ Nx and f(oX[p]) ⊆ f(Nx) ⊆ oY
⇒ ∃ oX[p] ∈ τX s.t. p ∈ oX[p] ⊆ f-1(oY)
⇒ f-1(oY) = ∪ oX[p] is open

Let BX be Bases and SX be Subbasis for topology τX
Let BY be Bases and SY be Subbasis for topology τY

CT2 ⇒ CT4
∀b ∈ BY
⇒ b ∈ τY [ Open set is a union of bases ]
⇒ f-1(b) ∈ τX

CT4 ⇒ CT2
∀o ∈ τY
⇒ o = ∪b[i] where b[i] ∈ BY
⇒ f-1(o) = f-1(∪b[i]) = ∪f-1(b[i])
⇒ f-1(o) ∈ τX [ By f-1(b[i]) ∈ τX and topology axiom T2 ]

CT4 ⇒ CT5
∀s ∈ SY
⇒ s ∈ BY [ Basis is a finite intersection of subbases ]
⇒ f-1(s) ∈ τX

CT5 ⇒ CT4
∀b ∈ BY
⇒ b = ∩s[i] where s[i] ∈ SY
⇒ f-1(b) = f-1(∩s[i]) = ∩f-1(s[i])
⇒ f-1(b) ∈ τX [ By f-1(s[i]) ∈ τX and topology axiom T3 ]

Consider the Unit Step Function f:R→R again in the previous example. Now, f-1((0.4, 0.6)) = {0}, (0.4, 0.6) is an open set in R but {0} is not an open set in R. Thus, f is not a continuous mapping.

The continuity of mapping f:X→Y depends on the topology defined on X and Y. The following example shows that when the topology changes, the mapping can be changed from a continuous one into a discontinuous one.

Let τ = {o|o = ∪(a,b)} and τ’ = {o|o = ∪[a,b)}. Both (R, τ) and (R, τ’) are topological spaces for R. τ is called Standard Topology on R and τ’ is called Lower Limit Topology on R.
Consider the following 2 identity functions f & g:
f:(R, τ)→(R, τ’) f(x) = x
g:(R, τ’)→(R, τ) g(x) = x

∀[a,b) ∈ τ’, f-1([a,b))=[a, b) ∉ τ [ By definition of τ ]
So, f is not a continuous mapping.

∀(a,b) ∈ τ, g-1((a,b))=(a,b)=∪[a’,b) where a’= a+(b-a)/(n+1) as n→∞
b-a’=b-[a+(b-a)/(n+1)]=(b-a)(1–1/(n+1)) > 0 [ By n>0 and b > a ]
⇒ [a’,b) ∈ τ’
⇒ (a,b)=∪[a’,b) ∈ τ’ [ By topology axiom T2]
So, g is a continuous mapping.

# Examples of Continuous Real Function

Let’s examine some elementary functions which are continuous. Standard Topology (R, τ) on R will be used in the proof.

Denote
B(a,e) = {x∈Rn| ∥x-a∥ < e}
f+: RxR → R
f+(x,y)=x+y

Proof-
For any open set o in (R, τ), ∀c∈o, there exist e>0 s.t. B(c,e) ⊆ o
Consider B((a,b), e/2) where f+(a,b) = c
∀(x,y) ∈ B((a,b),e/2)
f+(x,y)-c∥=∥x+y-a-b∥=∥x-a+y-b∥<∥x-a∥+∥y-b∥< e/2+e/2=e
⇒f+(x,y) ∈ B(c,e) ⊆ o
⇒f+(B((a,b), e/2)) ⊆ o
⇒B((a,b),e/2) ⊆ f+-1(o) ⇒f+-1(o) is open in (RxR, τ) ⇒ f+ is continuous

Multiplication function -
f*: RxR → R
f*(x,y)=x*y

Proof-
For any open set o in (R, τ), ∀c∈o, there exist e>0 s.t. B(c,e) ⊆ o
Consider B((a,b),e’) where f*(a,b)=c
∀(x,y) ∈ B((a,b),e’)
f*(x,y)-c∥=∥xy-ab∥= ∥(x-a)(y-b)+xb-ab+ay-ab∥ = ∥(x-a)(y-b)+(x-a)b+a(y-b)∥ < ∥x-a∥∥|y-b∥+|b|∥x-a∥+|a|∥y-b∥ < e’(e’+|a|+|b|)
So, let e’ = min(1, e/(1+|a|+|b|))
case 1: if e/(1+|a|+|b|) >= 1,
⇒ e’ = 1,
f*(x,y)-c∥ < (1+|a|+|b|) <= e
case 2: if e/(1+|a|+|b|) < 1,
⇒ e’=e/(1+|a|+|b|) < 1
f*(x,y)-c∥ < e’(1+|a|+|b|) = e
⇒f*(x,y) ∈ B(c,e) ⊆ o in both cases 1 & 2
⇒f*(B((a,b),e’)) ⊆ o
⇒B((a,b),e’) ⊆ f*-1(o) ⇒f*-1(o) is open in (RxR, τ)⇒ f* is continuous

Reciprocal function -
f/: R\{0} → R
f/(x)=1/x

Proof-
For any open set o in (R, τ), ∀c∈o, there exist e>0 s.t. B(c,e) ⊆ o
Consider B(a, e’) where f(a)=c
If a>0 ⇒|x-a|< e’ = a/2 ⇒ -a/2 < x-a <a/2 ⇒ a/2 < x < 3a/2 ⇒2/a > 1/x > 2/3a > 0
|1/x-1/a|=|x-a|/|x||a| < e’1/a*2/a = 2e’/a²
So, let e’ = min(a/2, (e/2)*a²)
case 1: if a/2 < e/2*a² ⇒1 < e*a, e’=a/2
⇒|f/(x)-c| = |1/x-1/a| < a/2*1/a*2/a=1/a < e
case 2: if a/2 >= e/2*a² ⇒ 1 >= e*a, e’=e/2*a²
⇒|f/(x)-c| = |1/x-1/a| < e/2*a²*1/a*2/a = e

If a<0 ⇒|x-a|< e’ = -a/2 ⇒ a/2 < x-a < -a/2 ⇒ 3a/2 < x < a/2 ⇒0 > 2/3a > 1/x > 2/a
|1/x-1/a|=|x-a|/|x||a| < e’/xa < e’*1/a*2/3a = 2e’/3a²
So let e’ = min(-a/2, (3e/2)*a²)
case 3: if -a/2 < (3e/2)*a² ⇒ -1 < 3e*a, e’ = -a/2
⇒|f/(x)-c| = |1/x-1/a| < -a/2*1/a*2/3a = -1/3a < e
case 4: if -a/2 >= (3e/2)*a² ⇒ -1 >= 3ea, e’ = (3e/2)*a²
⇒ |f/(x)-c| = |1/x-1/a| < (3e/2)*a²*1/a*2/3a = e
⇒ f/(x) ∈ B(c,e) ⊆ o in all cases 1–4
⇒ f/(B(a, e’)) ⊆ o
⇒ B(a, e’) ⊆ f-1(o)⇒f-1(o) is open in (R, τ) ⇒ f/ is continuous on R\{0}

Constant function -
f: R → R
f(x)=d where d ∈ R

Proof-
For any open set o in (R, τ),
If d ∈ o ⇒ f-1(o) = R
If d ∉ o ⇒ f-1(o) = ø
⇒ f-1(o) is open in (R, τ) ⇒ f is continuous

Vector valued function -
vf: X→Rn, f1: X→R, f2: X→R, … fn:X→R
vf(x) = (f1(x), f2(x), .. fn(x))
f1, f2..fn are all continuous functions

Proof-
For any open set o in (Rn, τ), ∀c∈o, there exist e>0 s.t. B(c,e) ⊆ o
Consider any a ∈ X where f(a)=c, f1(a)=c1, f2(a)=c2,..fn(a)=cn

f1,f2..fn are all continuous, so f1–1(B(c1,e’)), f2–1(B(c2,e’)).. fn–1(B(cn,e’)) are all open.

Let uc = f1–1(B(c1,e’)) ∩ f2–1(B(c2,e’)).. ∩ fn–1(B(cn,e’))
uc is open as f1–1(B(c1,e’)), f2–1(B(c2,e’)).. fn–1(B(cn,e’)) are all open. Since a ∈ u, u is not empty.

∀x ∈ uc ⇒
∥vf(x)-c∥=∥vf(x)-vf(a)∥=∥(f1(x),f2(x),..,fn(x)) — (f1(a),f2(a),..fn(a))∥
=∥(f1(x)-f1(a), f2(x)-f2(a),..,fn(x)-fn(a))∥
= √((f1(x)-c1)²+(f2(x)-c2)²+..+(fn(x)-cn)²) < √(ne’²)

Let e’=e/√n
vf(x)-c∥ < e
⇒ vf(x) ∈ B(c,e) ⊆ o
⇒ vf(uc) ⊆ o
⇒ vf is continuous at a => vf is continuous since a is an arbitrary element of X

Composition function -
f: X → Y, g: Y → Z
f and g are both continuous functions

g∘f: X → Z
g∘f(x)=g(f(x))

Proof-
∀ oZ ∈ τZ, ∃ oY=g-1(oZ) ∈ τY. [g is continuous]
⇒ ∃ oX=f-1(oY) ∈ τX where oX=f-1(g-1(oZ))=(g∘f)-1(oZ) [f is continuous]
⇒ g∘f is continuous

Restriction function -
f: X → Y
f is continuous
A ⊂ X equipped with subspace topology

f|A: A → Y
f|A(x) = { f(x)| x ∈ A }

Proof-
∀ oY ∈ τY, ∃ oX=f-1(oY) ∈ τX
⇒ ∃ oX ∩ A ∈ τA s.t. f|A(oX ∩ A) = f(oX ∩ A) ⊂ f(oX) = oY [ since A ⊂ X ]
⇒ f-1|A(oY) is open
⇒ f|A is continuous

# Conclusion

In this post, we have studied the definition of continuous mapping between Topological Spaces and how this definition is generalised from ε-δ definition.

In the coming next 2 articles, I will show you the concept of Closure and one application of topology theory for constructing continuous function — Urysohn’s lemma.

# Please give this post a clap if you like this post. Thanks!!

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"The fear of the LORD is the beginning of wisdom" 🙏 Seasoned Java Programmer. MSc Comp Sc, BSc Math (1st Hon), BSc Mech Eng