# General Topology — Part 2 (Continuous Function)

*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}**Addition 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/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**.

# Your feedback is highly appreciated and will help me to continue my articles.

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Reference:

“https://en.wikipedia.org/wiki/Topological_space”

“https://en.wikipedia.org/wiki/Open_set”

“https://en.wikipedia.org/wiki/Base_(topology)”

“https://en.wikipedia.org/wiki/Subbase”

“https://en.wikipedia.org/wiki/Subspace_topology”

“https://en.wikipedia.org/wiki/Continuous_function”

“https://en.wikipedia.org/wiki/Lower_limit_topology”

“Topology (2nd edition) by James Munkres”