When I was studying topology, I kinda disregarded the Quotient Topology part. It was short, but I didn't see much use of it inside the theory we were looking at during the course. Turns out I was right, but the conclusion - "so, disregard it" turned out to be quite wrong, since the knowledge is both aesthetic and useful.

Well, to begin with, I think it's a good idea to explain what is a Topological Space.  Well, consider you are in a metric space. The open sets of the space are sets $A$ such that each point of $A$ has a ball around it still inside of $A$. We will say any open set around a point is a neighbourhood of the point. Open sets have two interesting properties:

1 - Arbitrary union of open sets is still open.
2- Finite intersection of open sets is still open.

Turns out these two properties are enough to demonstrate a good amount of results. Just like when we leave $\mathbb{R}^n$ and go to metric spaces, having realized that (for a lot of purposes) all we need is the triangle inequality (plus some good conditions) to talk meaningfully of distance, we will abandon metric spaces, desiring to talk meaningfully of neighbourhoods (open sets). This generalization is, in my point of view, a good motivation by itself. But the fact is that topology has far-reaching uses in a lot of fields of mathematics, and helps us understand and formalize things like "a circle is a closed interval with its endpoints glued".

So, what is a Topological Space?

Definition: A Topological Space is a set $X$, together with a set $\tau$ of subsets of $X$, such that:

i) $\emptyset \in \tau$; $X \in \tau$
ii) Given a family $A_{\alpha}$ in $\tau$,   $\displaystyle \bigcup A_{\alpha} \in \tau$
iii) $A_1, A_2 \in \tau \Rightarrow A_1 \cap A_2 \in \tau$

The elements of $\tau$ are called open sets of the topology.

Examples:

i) Metric Space - A metric space is a topological space, where the open sets are precisely its open sets (note that the first "open sets" refers to "the open sets of the topology", whereas the second "open sets" refers to the "open set as defined in a metric space).
ii) Given a set $X$, the topology $\tau=\{\emptyset, X\}$ is called the trivial topology. It is easily verified that it is a topology.
iii) Given a set $X$, the topology $\tau=\mathcal{P} (X)$ is called discrete topology. It is also easily verified that it is a topology.

For an example that is new to first-readers of topology:

iv) Given a set $X$, consider $\tau=\{\text{subsets of } X \text{ that have finite complement, and the empty set}\} $. This is a topology, called the finite complement topology, and a simple-to-describe one. Moreover, it has interesting features. First, let's define a property that is kind of reasonable to expect:

Definition: We say the topological space $(X,\tau)$ is Hausdorff  if, for every two points $p_1, p_2$ of $X$, there exists a neighbourhood $U_1$ of $p_1$ and a neighbourhood $U_2$ of $p_2$ such that $U_1 \cap U_2=\emptyset$. 

Essentially, this says we can "properly" separate points. It is obvious a metric space is a Hausdorff topological space. In fact, every topological space should be Hausdorff... or not?

Take our example (iv): Consider the set $\mathbb{Z}$ with the finite complement topology. Take the points $1$ and $2$. A neighbourhood $U_1$ of $1$ is, for example, $\{1,2,6,7,8,9,10,...\}$ and a neighbourhood $U_2$ of $2$ is, for example, $\{2,3,4,1210,1430,1431,1432,1433,1434,...\}$. But it doesn't matter what neighbourhoods of $1$ and $2$ you use, since they always have finite complement, there is a number $N_1$ such that any $n>N_1$ will be in $U_1$. Likewise, there is a number $N_2$ such that any $n>N_2$ will be in $U_2$. So, the intersection will always be non-empty ($\max\{N_1, N_2\}+1$ will be in the intersection, for example). Therefore, $\mathbb{Z}$ with the finite complement topology is not Hausdorff.

Well, now that we know what is a topological space, let's introduce one of the most fundamental concepts of Topology: Continuous Functions. As we will see, continuous functions with a continuous inverse will behave quite analogous to what a "isomorphism" is in linear algebra.

So, let's begin with the analysis definition of continuity:

Definition 1 of continuity: $f:X\rightarrow Y$ is said to be continuous at $x_0$ if $\forall \epsilon > 0 \exists \delta > 0 : d(x,x_0)<\delta \Rightarrow d(f(x),f(x_0))<\epsilon$.

There is, in this definition, an explicit use of distance. Here comes one that is easy to see to be equivalent, but quite more appealing:

Definition 2 of continuity: $f:X\rightarrow Y$ is said to be continuous at $x_0$ if for all neighbourhood $V$ of $f(x_0)$, there exists a neighbourhood $U$ of $x_0$ that is taken inside $V$ by $f$.

In fact, this definition is automatically generalized to topological spaces, where there is no distance. Now, we have the following proposition:

Proposition: A function $f:X \rightarrow Y$ is continuous if and only if, for every open set $V$ of $Y$, $f^{-1}(V)$ is open. 



Note that this can be rephrased more compactly with: Inverse image of open is open.


Demonstration: Let's prove $\Rightarrow$:
Take an open set $V$ of $Y$. We have to prove $f^{-1}(V)$ is open. So, take a point $p$ in $f^{-1}(V)$. By continuity, there is a neighbourhood $U$ of $p$ that goes into $V$. So, this implies $U \subset f^{-1}(V)$. Therefore, $p$ is an interior point. Since $p$ is arbitrary, $f^{-1}(V)$ is open.
For $\Leftarrow$, 
let $p$ be a point of X. We will prove $f$ is continuous at $p$. Let $V$ be a neighbourhood of $f(p)$. By hypothesis, $f^{-1}(V)$ is open. Hence, there is a neighbourhood $U$ of $p$ in $f^{-1}(V)$. So, this neighbourhood will be taken to $V$. Therefore, $f$ is continuous at $p$. $\square$

Turns out that, for topological purposes, this property of continuity is so important that it is common to define  "continuous" as "pre-image of open is open". In fact, if a continuous function is bijective and has a continuous inverse (a homeomorphism), it is easy to see that it will take open sets from one topology to open sets of another topology. Hence, homeomorphisms "preserve the topology" in some sense. If there is a homeomorphism between two topological spaces, we can say they are topologically equivalent.

This "Pt.1" serves as an introduction to "Pt.2", where I'll talk about quotient spaces, and adjunction spaces.

This post is essentially an adaptation to an answer I gave to a question on MathSE. In there, I gave another approach that may be worth checking: response in MathSE. 

Initial observation: I'll assume that it is well-understood what is a number raised to a rational power.

Firstly, there are a lot of "motivational" approaches to the study of the exponential function. I'll adopt a purely mathematical motivation (specifically, algebraic) that is essentially in the nucleus of almost all others. 

We know that there are two canonical algebraic structures in $ \mathbb{R}$: addition and multiplication. Formally, we consider the groups: $ (\mathbb{R},+)$ and $(\mathbb{R}_{> 0},\times)$. The question we ask is: 

• Is there an isomorphism between these two structures?

In other words, are they algebraically the same? (That, in practical terms, means that "doing caculations in one of them is the same as in the other. That is equivalent to saying that you would be able to multiply two numbers essentially by doing an addition and transforming things. Transforming multiplication into addition should already be enough to justify a study of such function) 

Now, let's try to extract some properties of what a function of this type, which we will call $\xi : (\mathbb{R},+) \rightarrow (\mathbb{R}_{> 0},\times)$ should have. Firstly, she should satisfy, by its definition, the following equation:

$\xi(x+y)=\xi(x).\xi(y)$

Since $\xi(0)=\xi(0+0)=\xi(0)\xi(0)$, $\xi(0)=1$.

Now, define $k:= \xi(1)$. What we will show now is that, with the value of $k$, we have the value of $\xi$ in all rationals.

1. Firstly, if $n \in \mathbb{N}$, we have$\xi(n)=\xi(1)...\xi(1)$ $ n$ times=$\xi(1)^n$
2. If $-n$ is a negative integer, we have $ 1=\xi(0)=\xi(n-n)=\xi(n)\xi(-n)$, which implies: $\displaystyle \xi(-n)=\frac{1}{\xi(n)}=\frac{1}{\xi(1)^n}=\xi(1)^{-n}$
3. If $a=\frac{p}{q} \in \mathbb{Q}$, $ \xi(p)=\xi(q.\frac{p}{q})=\xi(\frac{p}{q}+\frac{p}{q}+...+\frac{p}{q})$ $ q$ times $ =\xi(p/q)^q$. So, $ \xi(\frac{p}{q})=\xi(p)^{1/q}=(\xi(1)^p)^{1/q}$

We conclude that if we have the value of the function in $1$, we have its value in all rationals. But we want in all real numbers. So, we will try to find a function that is of $\xi$'s type... maybe we can find an analytic one? Let's investigate:

We want to find a $\displaystyle \xi_1 (x)=\sum_{n=0}^{\infty}a_nx^n$ that is of the "type" of $ \xi$, meaning: $\xi_1 (x+y)=\xi_1(x)\xi_1(y)$. Do:

$ \displaystyle \xi_1(x+y)=\sum_{n=0}^{\infty}a_n(x+y)^n=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_n\frac{n! x^ky^{n-k}}{k!(n-k)!}=\sum_{n=0}^{\infty}a_nn!\sum_{k=0}^{n}\frac{ x^ky^{n-k}}{k!(n-k)!}$ , by the Binomial Theorem.

Notice that the last equality is BEGGING for the Cauchy Product Formula. If $ a_nn!$ was $1$... the last side would be $\displaystyle \left(\sum_{k=0}^{\infty}\frac{x^k}{k!}\right)\left(\sum_{m=0}^{\infty}\frac{y^m}{m!}\right)$. But we have the power to choose who the $a_n$ are! We are searching a candidate to the series, after all. So, take $a_nn!=1$, which implies $ \displaystyle a_n=\frac{1}{n!}$. Now, stare at our luck! Since $\displaystyle a_n=\frac{1}{n!}$, what we have in the last equality is precisely the product $ \xi_1(x).\xi_1(y)$.

So, the function $\displaystyle \xi_1(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ satisfies all we want. (The series is well defined for all x real, which is easily verifiable by the ratio test. The series is obviously greater than zero is x is greater than 0, and the fact that $\xi_1(x).\xi_1(-x)=\xi_1(0)=1$ implies the series is greater than zero for all x real. Other observations also imply that it is bijective, since it is surjective (follows from the fact that it goes to infinity when x goes to infinity, to zero when x goes to -infinity, and it is continuous) and injective (because it is strictly increasing)

As we saw, it is enough knowing $\xi_1$ in $1$ to know its value in all rationals. But we know what it is in $1$:

$\displaystyle \xi_1(1)=\sum_{n=0}^{\infty}\frac{1}{n!}$

We will call this amazing number $e$.

So, $\xi_1(r)= e^r$ if $r$ is rational, in the usual sense we have for "a number raised to a rational power" (this follows from (1), (2) and (3)). Note that we already have an answer to our initial problem: we found a function $\xi_1$  which is our isomorphism between the groups. But it may be worth a while to try and say a few more things about it.

Since $\xi_1(r)=e^r$ if $r$ is rational, we define :

$e^x:=\xi_1(x)$

when $x$ is real, and this satisfies what we would wait of $e^x$ by previous observations. OK then, we found one function that extends what we call "exponentiation" of rational powers to real powers. But this function is analytic. Is there another less smooth function, for example $ C^{\infty}$ or even only continuous that could extend exponentiation? The answer is not, since this function would coincide in rational points, and the rationals are dense in $\mathbb{R} $ (Two continuous functions that coincide in a dense subset coincide in all $\mathbb{R}$). Then, this extension is unique if we require it to be continuous. This is enough to call  $\xi_1(x)$ "the exponential", and have no regrets for calling it $e^x$.

Now, since $e^x$ is bijective, there is an inverse, which we will call $ln(x)$. This allows us to define, for an arbitraty $ a >0 $ :

$a^x:=e^{ln(a).x}$

Since, by definition $ a^1=a$ e $ a^{(x+y)}=a^x.a^y$, it follows from the same considerations as before that this function is what deserves to be called "raising $a$ to a real power".

Thus, we have:

$ (e^x) ^y=e^{(ln(e^x)).y}=e^{(x.y)}$

But going back to our previous question: what if we forget the constraint of a continuous extension: is there another function that coincides with $e^x$ on the rationals?

Yes, but we will need the following theorem to "show" a function of this kind:

Theorem: Every vector space has a basis.

Now, consider $ \mathbb{R}$ as a vector space over $ \mathbb{Q}$, and take a basis $ \{x_i\}_{i \in \Lambda}$ that has $1$. Define:

$f(1):=e$

$ f(x_i):=e^{x_i/2}$ if $ x_i \neq 1$

Since we defined the function on the basis and we want that $f(x+y)=f(x).f(y)$, we define:

$ f(x):=f(x_1)^{\lambda_1}f(x_2)^{\lambda_2}...f(x_n)^{\lambda_n}$,

where $ x=\lambda_1x_1+\lambda_2x_2+...+\lambda_nx_n$ is the representation of $x$ as a linear combination of elements of the basis (keep in mind that $\lambda_i$ are rationals!), and everything is well defined since$\{x_i\}$ is a basis. The fact that this function satisfies $f(x+y)=f(x).f(y)$ follows from definition, and the non-continuity of the function follows from the fact that it coincides with the exponentiation on the rationals but is not the exponentiation on the reals (she takes different values on $ x_i$, for example).

We see that requiring continuity is not only very reasonable, but also necessary. Otherwise, we would have a lot of "non-constructive exponentials" to consider.

This blog will be about Mathematics in general. Probabily, I'll talk about the following subjects:

- Things I'm studying. This way, I help myself to grasp new concepts etc.
- Things I've already studied. In this case, I'll try to talk about things/conclusions/points of view that I arrivied in my study independently: the subjects will be adressed a little differently than books etc (since it comes from my head)
- Mathematical Curiosities
- Phylosophical aspects of Mathematics
- My thoughts about certain things (that will always be about Mathematics. I don't intend doing any form of political or ideological discussion in this space.

Any mistakes, suggestion, complaints etc, please tell me.

Kind Regards,
Aloizio