continuous functions

Continuous Function

Let

X, Y

be topological spaces, a function

f : X \to Y

is said to be continuous iff for each open subset

V

Y

f^{- 1} (V)

is an open set of

X

Continuous iff Every Basis Element is Open

Suppose that

f : X \to Y

is a function and

T_{Y}

is generated by a basis

B

, then

f

is continuous if every basis element

f^{- 1} (B)

is open in

Y

Let $U$ be open in $Y$ , we want to prove that $f^{- 1} (U)$ is an open set. Since $U$ is open then we know that $U = ⋃ C$ where $C \subseteq B$ .

By noting that $f^{- 1} (U) = f^{- 1} (⋃ C)$ $=$ $⋃ {f^{- 1} (C) : C \in C}$ , since $C \subseteq B$ , then every element $C$ is a basis element, and thus $f^{- 1} (C)$ is an open set by our assumption, therefore as it is an arbitrary union of open sets, we can see that $f^{- 1} (U)$ is an open set, as needed

The Set of Points Where a Function Is Larger Than the Other Is Open

Let

Y

be a set with the order topology, and

f, g : X \to Y

then the set

A = {x : f (x) > g (x)}

is open in

X

Let

p \in A

now let's an open set around

x

contained in

A

. Since

p \in A

then we know that

f (p) > g (p)

therefore since

Y

is hausdorff then we know that there are disjoint neiborhoods

U, V

g (p), f (p)

respectively such that

\forall u, v \in U, V

we have

u < v

. Since

f, g

are continuous we also knot that

S = g^{- 1} (U) \cap f^{- 1} (V)

is open in

X

, if

s \in S

then we know that

g (s) \in U

and that

f (s) \in V

and so

g (s) < f (s)

therefore

s \in A

so that

S \subseteq A

and therefore it is open.

Note that as a corollary to this we know that the set ${x \in X : f (x) \leq g (x)}$ is closed.

The Min Function of Two Functions Is Continuous

Suppose

Y

is the order topology and that

f, g : X \to Y

are continuous, then the function

h : X \to Y

defined as

h (x) = min {f (x), g (x)}

is continuous

Let $A = {x \in X : f (x) \leq g (x)}$ and $B = {x \in X : g (x) \leq f (x)}$ which we know are closed, and thus $A \cap B$ is also closed, moreover we know that $A \cap B = {x \in X : f (x) = g (x)}$ but also that $A \cup B = X$ in order to use the pasting lemma we need to restrict our functions $f, g$ to $f^{'}, g^{'}$ so that $f^{'} : A \to Y$ and $f^{'} : B \to Y$ (which we know are continuous as a restriction of a continuous function).

With these functions we can notice that $h (x) = min {f (x), g (x)} = {\begin{matrix} f^{'} (x) & if x \in A \\ g^{'} (x) & if x \in B \end{matrix}$ thus by invoking the pasting lemma we conclude that $h (x)$ is a continuous function which is what we wanted to show.

Finite to One

Given a function

f : X \to Y

we say that it is finite to one diff for every

y \in Y

| f^{- 1} ({y}) | < \infty

Continuous Identity Implies One Topology Is Finer Than the Other

Suppose that

T_{1}, T_{2}

are both topologies on

X

then identity

id : (X, T_{1}) \to (X, T_{2})

is continuous, then

T_{1} \supseteq T_{2}

Suppose that

U \in T_{2}

, then since

{id}^{- 1} = id

then we know that

{id}^{- 1} (U) = U

is open in

T_{1}

since

id

is continuous, therefore we are done

Two Topologies Are Equal If the Identity Is a Homeomorphism

Suppose that

T_{1}, T_{2}

are both topologies on

X

then identity

id : (X, T_{1}) \to (X, T_{2})

is a homeomorphism then

T_{1} = T_{2}

A Function Is Continuous When Taken With the Finite Complement Topology Iff It Is Constant or Finite One to One

Suppose that

f : X \to Y

where

X, Y

are taken with the finite complement topology. Then

f

is continuous iff it is constant or finite to one.

Suppose that $f$ is continuous, then it's either true or false that $f$ is constant, if $f$ is constant then the proof is done, so let's suppose that it is not constant, so we must prove it is finite one to one, let $y \in Y$ , since $Y ⧵ (Y ⧵ {y}) = {y}$ this shows that $Y ⧵ {y}$ is open in $Y$ therefore $f^{- 1} (Y ⧵ {y})$ which equals $f^{- 1} (Y) ⧵ f^{- 1} ({y})$ but also equals $X ⧵ f^{- 1} ({y})$ is an open set. Since it is, this implies that $f^{- 1} ({y})$ is either finite or all of $X$ , but since $f$ is not constant, then it cannot be all of $X$ therefore we know that $f^{- 1} ({y})$ is finite, showing that $f$ is finite to one.

Now if $f$ is constant, then suppose $f (x) = c$ for any $x \in X$ therefore if we're given an open $V$ in $Y$ then if $c \in V$ then $f^{- 1} (V) = X$ and if not then $f^{- 1} (V) = \emptyset$ , in either case the inverse image is open. On the other hand if $f$ is not constant, we shall prove $f$ is finite to one, then still supposing that $V$ was open in $Y$ then that means that $Y ⧵ V$ is finite or $V = \emptyset$ (But in that case $f^{- 1} (\emptyset) = \emptyset$ which is open in $X$ ), so supposing that that difference was finite, then we know that $f^{- 1} (Y ⧵ V) = ⋃_{p \in Y ⧵ V} f^{- 1} (p)$ which is a finite union of finite sets and thus finite, but we also know that $f^{- 1} (Y ⧵ V) = X ⧵ f^{- 1} (V)$ therefore $f^{- 1} (V)$ is open in $X$ which is exactly what we wanted to prove.

Continuous iff Every Subbasis Element is Open

Suppose that

f : X \to Y

is a function and

T_{Y}

is generated by a subbasis

S

, then

f

is continuous if every subbasis element

f^{- 1} (S)

is open in

X

Since $T_{Y}$ is generated by $S$ , this means that $T_{Y} = T_{ℬ_{S}}$ where $B_{S}$ is the basis generated by $S$ therefore to verify that this function is continuous all we have to do is check that every basis element $B \in B_{S}$ has the property that $f^{- 1} (B)$ is open with respect to $X$

We know $B = S_{1} \cap S_{2} \cap \dots \cap S_{k}$ for some $k \in N_{1}$ , therefore we want to know if $f^{- 1} (S_{1} \cap S_{2} \cap \dots \cap S_{k})$ is open, since we know that the intersection factors through the inverse image then we can re-write this as $⋂_{i = 1}^{k} f^{- 1} (S_{i})$ .

Our initial assumption was that given any $S \in S$ that $f^{- 1} (S)$ was open, thus specifically for our $S_{1}, \dots, S_{k}$ their inverse images should also be open, therefore $⋂_{i = 1}^{k} f^{- 1} (S_{i})$ is a finite intersection of open sets of $X$ thus it is open, which shows that $f$ is continuous.

Continuous Implies Image Closure Expansion

Suppose that

f

is continuous, then given any subset

A

X

, we have that

f (\overset{―}{A}) \subseteq \overset{―}{f (A)}

Suppose that $f$ is continuous, and suppose that $A \subseteq X$ , we want to show that $f (\overset{―}{A}) \subseteq \overset{―}{f (A)}$ , we start by assuming that $p \in f (\overset{―}{A})$ , which means that $p = f (a)$ where $a \in \overset{―}{A}$ , we now want to prove that $p \in \overset{―}{f (A)}$ , one way of going about this would be to show that every neighborhood of $p$ intersects $f (A)$ , which we will do.

So let $V$ be a neighborhood of $p$ , by definition we know $V \subseteq Y$ , then we know that $f^{- 1} (V)$ is an open set since $f$ is continuous, since $p = f (a)$ , then $V$ is also a neighborhood of $f (a)$ , therefore clearly $f (a) \in V$ , that is $a \in f^{- 1} (V)$ .

We assumed that $a \in \overset{―}{A}$ , which is equivalent to every neighborhood of $a$ intersecting $A$ , we know that $f^{- 1} (V)$ is an open set containing $a$ , which makes it a neighborhood of $a$ , and thus it must intersect with $A$ at some point $b$ , then $f (b) \in V$ , at the same time we know that $b \in A$ as well, so that $f (b) \in f (A)$ , thus $f (b) \in V \cap f (A)$ showing that the neighborhood intersects $f (A)$ , therefore this shows that $p \in \overset{―}{f (A)}$ , as needed.

Closure Expansion Implies Closed Inverse Image

Suppose that for every subset

A \subseteq X

, we have

f (\overset{―}{A}) \subseteq \overset{―}{f (A)}

, then for every closed set

B

Y

f^{- 1} (B)

is closed in

X

Let $B \subseteq Y$ be a closed set of $Y$ , we want to prove that $A = f^{- 1} (B)$ is a closed set of $X$ , which is equivalent to $A = \overset{―}{A}$ , therefore we will prove this instead.

We already know $A \subseteq \overset{―}{A}$ , so all we need to show is that $\overset{―}{A} \subseteq A$ , so let $x \in \overset{―}{A}$ , and we want to prove that $x \in A = f^{- 1} (B)$ , in other words we want to show that $f (x) \in B$ .

Before we continue, we should realize that $A$ is a subset of $X$ , therefore our assumption applies, which tells us $f (\overset{―}{A}) = \overset{―}{f (A)}$ , since $A = f^{- 1} (B)$ , we can conclude that $f (\overset{―}{A}) \subseteq B$ .

Since $x \in \overset{―}{A}$ , then $f (x) \in f (\overset{―}{A})$ , but then by our previous paragraph, we know that $f (x) \in B$ , as needed.

Closed Inverse Image Implies Continuous

Suppose that for every closed

B

Y

f^{- 1} (B)

is closed in

X

, then

f

is continuous

Let $V$ be an open set of $Y$ , we want to show that $f^{- 1} (V)$ is an open set of $X$ , set $B = Y ⧵ V$ .

Since $V$ was open, then $B$ is closed, and thus $f^{- 1} (B)$ is a closed set. But note this: $f^{- 1} (B) & = f^{- 1} (Y ⧵ V) & = f^{- 1} (Y) ⧵ f^{- 1} (V) \tag $ α $ & = X ⧵ f^{- 1} (V) \tag $ β $$

Where we've used the facts:

$α$
$β$

The conclusion from the above was that $f^{- 1} (B) = X ⧵ f^{- 1} (V)$ , from our assumption we knew that $f^{- 1} (V)$ was a closed, set, which makes $f^{- 1} (B)$ an open set, which is what we set out to prove.

Continuous iff Every Neighborhood Contains the Image of another Neighborhood

f

is continuous iff for every

x \in X

and neighborhood

V

f (x)

, there is a neighborhood

U

x

such that

f (U) \subseteq V

$⟹$ Let $x \in X$ and suppose $V$ is a neighborhood of $f (x)$ , then $f^{- 1} (V)$ is an open set since $f$ was assumed continuous, it also must contain $x$ , this is because $f (x) \in V$ , making $f^{- 1} (V)$ a neighborhood of $x$ , we also know $f (f^{- 1} (V)) \subseteq V$ , as needed.

$⟸$ Suppose $V$ is open in $Y$ , if $f^{- 1} (V) = \emptyset$ , we are done, thus we can find some element $x \in f^{- 1} (V)$ , making $V$ a neighborhood of $f (x)$ , thus by assumption, we get some neighborhood $U_{x}$ such that $f (U_{x}) \subseteq V$ , this by definition means that for any $p \in U_{x}$ , we know $f (p) \in V$ , or that $p \in f^{- 1} (V)$ , which shows that $U_{x} \subseteq f^{- 1} (V)$ .

Since this is true for every point $x \in f^{- 1} (V)$ , since $⋃_{x \in f^{- 1} (V)} U_{x} = f^{- 1} (V)$ , then $f^{- 1} (V)$ can be written as an arbitrary union of open sets, and thus it is open, as needed.

Continuous, Image Closure Expansion, Closed Inverse Image, Nested Neighborhoods Equivalence

Let

X, Y

be topological spaces, let

f : X \to Y

, then the following are equivalent:

$f$ is continuous
For every subset $A$ of $X$ , $f (\overset{―}{A}) \subseteq \overset{―}{f (A)}$
For every closed set $B$ of $Y$ , $f^{- 1} (B)$ is closed in $X$
For each $x \in X$ and each neighborhood $V$ of $f (x)$ , there is a neighborhood $U$ of $x$ such that $f (U) \subseteq V$

We've shown that $1 ⟹ 2$ , $2 ⟹ 3$ , $3 ⟹ 4$ and $4 ⟺ 1$ , this induces directed graph that is weakly connected and has a cycle, therefore it is connected, which means that all statements are equivalent.

Homeomorphism

Let

X, Y

be topological spaces, and let

f : X \to Y

be a bijection. If the function

f

and it's inverse function

f^{- 1} : Y \to X

are continuous, then

f

is called a homeomorphism

Homeomorphism Set and it's Image are Open Equivalence

Suppose that

f

is a bijection, then

f

is a homeomorphism iff the following is true:

f (U)

is open in

Y

iff

U

is open in

X

Suppose that $f$ is a homeomorphism, and that $U$ is a subset of $X$ , assume $f (U)$ is open in $Y$ , we'll prove that $U$ is open in $X$ , since $f$ is continuous, then we know that $f^{- 1} (f (U))$ is an open set of $X$ , since $f$ is a bijection, this means it's injective and thus $f^{- 1} (f (U)) = U$ as needed.

Now suppose that $U$ is open in $X$ , since we know that $f^{- 1}$ is continuous this means that ${(f^{- 1})}^{- 1} (U)$ is an open set of $Y$ , we've shown that ${(f^{- 1})}^{- 1} (U) = f (U)$ , thus, we've just shown that $f (U)$ is an open set of $Y$ , this concludes

At this point we've proven the main rightward implication. The other direction is quite a lot easier. We have to show that both $f$ and $f^{- 1}$ are continuous.

Suppose that $V$ is an open set of $Y$ , we want to show that $f^{- 1} (V)$ is an open set of $X$ , which is true iff $f (f^{- 1} (V))$ is an open set of $Y$ , since $f$ is surjective, then $f (f^{- 1} (V)) = V$ making it an open set of $Y$ , which shows that $f^{- 1} (V)$ is an open set of $X$

Now suppose that $U$ is an open set of $X$ , our goal will be to show that ${(f^{- 1})}^{- 1} (U)$ is an open set of $Y$ , as seen previously ${(f^{- 1})}^{- 1} (U) = f (U)$ , and since $U$ was open in $X$ , by our assmption we know that $f (U)$ is open in $Y$ , this concludes the proof.

Imbedding

Suppose that

f : X \to Y

is an injective continuous function, with

X

and

Y

topological spaces. By setting

Z = f (X)

as a subspace of

Y

, if

f^{'} : X \to Z

obtained by restricting the range of

f

is a homeomorphism, then we say that the original

f

is an imbedding of

X

Y

Constant Functions are Continuous

Let

X, Y

be topological spaces and suppose that

p \in Y

, if

f : X \to Y

is constant with value

p

, then

f

is continuous

Suppose that

V \subseteq Y

, if

p \in V

, then

f^{- 1} (V) = X

, otherwise

p \notin V

f^{- 1} (V) = \emptyset

, in either case we have an open set of

X

Inclusion is Continuous

Suppose that

A

is a subspace of

X

, then the inclusion function

ι : A \to X

is continuous.

Let

U

be an open set of

X

, then

ι^{- 1} (U) = U \cap A

, which by definition is open in the subspace topology of

A

as needed.

Compositions of Continuous Functions are Continuous

Suppose that

f : X \to Y

and

g : Y \to Z

are continuous then the map

g \circ f : X \to Z

is continuous

Let $U$ be open in $Z$ , we want to prove that ${(g \circ f)}^{- 1} (U)$ is an open set of $X$ .

To start if we look at ${(g \circ f)}^{- 1} (U)$ , we know this is equal to $f^{- 1} (g^{- 1} (U))$ , thus since $g$ is continuous we know that $S : = g^{- 1} (U)$ is an open set of $Y$ and then $f^{- 1} (S)$ is open in $X$ since $f$ was continuous, this shows that $g \circ f$ is continuous

Domain Restriction is Continuous

f : X \to Y

is continuous and if

A

is a subspace of

X

, then the restricted function

f ↾ A : A \to Y

is continuous

We note that

f ↾ A = f \circ ι

, therefore as the composition of two continuous functions, we know that

f ↾ A

is continuous

Restricting the Range maintains Continuity

Suppose that

f : X \to Y

is continuous. If

Z

is a subspace of

Y

containing

f (X)

, then the function

g : X \to Z

obtained by restricting the range is continuous

To show continuity we start by assuming that

B

is open in

Z

, therefore

B = Z \cap U

where

U

is open in

Y

Pasting

Let

X = A \cup B

where

A, B

are closed in

X

. Let

f : A \to Y

and

g : B \to Y

be continuous. If

f (x) = g (x)

for every

x \in A \cap B

then

f

and

g

combine to give a continuous function

h : X \to Y

defined as

h (x) = {\begin{matrix} f (x) & if x \in A \\ g (x) & if x \in B \end{matrix}

Let

C

be a closed subset of

Y

then we know that

h^{- 1} (C) = f^{- 1} (C) \cup g^{- 1} (C)

but since

f, g

were continuous then their inverse images are also closed, therefore as

h^{- 1} (C)

is a union of two closed sets we deduce that it is closed as well showing that

h

is continuous.

Given a Convergent Sequence Then the Continuous Image Also Converges

Let

f : X \to Y

be continuous, and suppose that

x_{n} \to x

then

f (x_{n}) \to f (x)

🏗️ ΘρϵηΠατπ🚧 (under construction)

🏗️ $Θ ρ ϵ η Π α τ π$ 🚧 (under construction)