Topological space (draft)

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This is my note of the video It’s not a full guide of topology.

I wanted to know

  • what is continuity
  • the meaning of “topology”

Topological space

Let $M$ be a set. Then, a choice $\mathcal{O}\subseteq\mathcal{P}(M)$ is called a topology on $M$ if

  1. $\phi\in\mathcal{O}$ and $M\in\mathcal{O}$,
  2. $O_i, O_j \in\mathcal{O} \Rightarrow\bigcap\{O_i, O_j\}\in\mathcal{O}$, and
  3. $\displaystyle O_i\in\mathcal{O} \Rightarrow \bigcup_i O_i\in\mathcal{O}$.

The pair $(M, \mathcal{O})$ is then called a topological space. An elements of a topology is called an open set.

Note that,

  • The union could be countable infinite.
  • For a set $M$, there could be a lot of topologies.
$|M|$The number of topology

For any set $M$, $\mathcal{O}=\{\phi,M\}$ is called a chaotic topology. (“chaotic” as the antonym of “ordered”.)

For any set $M$, $\mathcal{O}=\mathcal{P}(M)$ is called a discrete topology.

In a simple English

When a set $M$ is given, you can group the elements. The way of grouping is defined by the subsets of $\mathcal{P}(M)$. If you group the subsets in a special manner (to comply the three conditions above), the choice of subsets is called topology.

Standard topology

Note that, we can define the standard topology for real numbers $M=\mathbb{R}$, and after that we can extend it to $\mathbb{R}^{d}$.

Open ball

For a set $M=\mathbb{R}^{d}$, $\forall x \in \mathbb{R}, \forall r\in\mathbb{R}^+$, define an open ball of radius $r$ around point $x$ as follows:

$$ B_r(x):= \left\{y\in\mathbb{R}^{d} \middle| \sqrt{\sum_{i=1}^{d}\left(y_i-x_i\right)^2} < r \right\} $$

We chose 2-norm, but even you chose $2n$-norm, the structure of topology doesn’t matter(?)

Standard topology

For a set $M=\mathbb{R}^{d}$, we can define the standard topology $\mathcal{O}_{\mathrm{std}\ \mathbb{R}^{d}}$ (or simpler $\mathcal{O}_{\mathrm{std}}$) as follows:

$$ O\in\mathcal{O}_{\mathrm{std}\ \mathbb{R}^{d}} :\Leftrightarrow \forall p\in O: \exists r\in\mathbb{R}^+: B_r(p)\subseteq O $$

The pair $\left(\mathbb{R}^{d}, \mathcal{O}_{\mathrm{std}\ \mathbb{R}^{d}}\right)$ is a topological space.


First, prove that empty set $\phi$ and the whole set $\mathbb{R}^{d}$ are in $\mathcal{O}_{\mathrm{std}\ \mathbb{R}^{d}}$.

For empty set, check the definition

$$ \phi\in\mathcal{O}_{\mathrm{std}\ \mathbb{R}^{d}} \Leftrightarrow \forall p\in \phi: \exists r\in\mathbb{R}^+: B_r(p)\subseteq \phi $$

Change the form of RHS as follows:

$$ \forall p: p\in\phi \Rightarrow\exists r\in\mathbb{R}^+: B_r(p)\subseteq \phi $$

$p\in\phi$ is always false, so by ex falso quodlibet, the proposition is true.

For the entire set $M$:

$$ \forall p\in \mathbb{R}^{d}: \exists r\in\mathbb{R}^+: B_r(p)\subseteq \mathbb{R}^{d} $$

This proposition is true by definition of subset. (<- need to be more presice?)

Second, check the intersection condition:

Let $O_i, O_j \in \mathcal{O}_{\mathrm{std}}$. Chose any point $p\in O_i\cap O_j$. By definition of open sets, there exist two real number $r$ and $s$ s.t.:

$$ \forall p\in O_i: \exists r\in\mathbb{R}^+: B_r(p)\subseteq O_i \\\ \forall p\in O_j: \exists s\in\mathbb{R}^+: B_s(p)\subseteq O_j $$

Then, $B_{\min(r,s)}(p) \in O_i$ and $B_{\min(r,s)}(p) \in O_j$. Hence $B_{\min(r,s)}(p)\subseteq O_i\cap O_j$. This is true for any point $p\in O_i\cap O_j$, so $O_i\cap O_j \in \mathcal{O}_{\mathrm{std}}$.

Lastly, check the union condition:

Let $O=\{O_i| O_i\in\mathcal{O}_{\mathrm{std}}\}$ and randomply pick a set $O_x\in O$. By definition, $O_x\in\mathcal{O}_{\mathrm{std}}$, so the following holds:

$$ \forall p\in O_x: \exists r\in\mathbb{R}^+: B_r(p)\subseteq O_x \subseteq\bigcup_i O_i $$

$x$ is chosed randomly, so this proposition holds for all $x$. Hence,

$$ \forall p\in \bigcup_i O_i: \exists r\in\mathbb{R}^+: B_r(p)\subseteq\bigcup_i O_i $$

From those three condition, $\mathcal{O}_{\mathrm{std}}$ is indeed a topology. $\blacksquare$

Induced topology

Let $(M,\mathcal{O})$ be a topological space and $N\subseteq M$. Then,

$$ \mathcal{O}\vert_N := \left\{O\bigcap N\ \middle|\ O\in\mathcal{O} \right\} \subseteq \mathcal{P}(N) $$

is actually a topology on $N$, and we call it as the induced topology (or subspace topology, relative topology, trace topology).

You an easily prove by checking the three condition of a topological space.

As an example, consider $(\mathbb{R}, \mathcal{O}_{\mathrm{std}})$ and check if $(0,1]$ is open set. $(0,1]$ is obviously not an open set in the standard topology, but we can make it as open set by inducing another topology with $N=[-1,1]$, because $(0,1] = (0,2)\cap[-1,1] = (0,2)\cap N$. Hence $(0,1]$ is an open set of the topology $\mathcal{O}_{\mathrm{std}}\vert_N$.

Quotient topology

Rough sketch

In the topology theory, you would meet some cut&glue operations. This is related to a quotient topology. A quotient topological space has two structure, topology and equivalence relation. In general, this two structure is an independent concept. An equivalence relation devides/groups a set into disjoint parts, where as toplology not only groups a set but also has a concept of “union of group”, which can’t be introduced by equivalence relations.

Suppose we cut a flexible paper, which can be regarded as a part of $\mathbb{R}^2$ with the standard topology, and glue its edges so that we make some 3D structure. When you cut a paper, you define an equivalence relation (you can imagine translational symmetry). During glueing the paper, you implicitly defined which edge should be connected to which edge. Finally, the standard topology helps to glue the edges smoothly.


Let $(M,\mathcal{O})$ be a topological space. Define a quotient map associated with $\sim$ as follows:

$$ \begin{align} q:M&\rightarrow X/\sim \\\ x&\mapsto [x] \end{align} $$

With a quotient map, we can define a quotion topology as follows:

$$ \mathcal{O}_{M/\sim} := \left\{ U \in M\ \middle|\ q^{-1}[U] \in M \right\} $$

where $q^{-1}[U]$ is a preimage of $U$ under $q$.

Closed set

Do you think the negation of “open” is “closed”? In terms topology, it isn’t.

Let $(M.\mathcal{O})$ be a topological space. A set $C\subseteq M$ called closed if $M\backslash C$ is open.

In general, a subset of topological space can be:

  1. open
  2. closed
  3. open and closed
  4. open and not closed
  5. not open and closed
  6. not open and not closed

For example, the empty set $\varnothing$ is closed because $\varnothing = M\backslash M$, but the empty set is open set by definition. Hence the empty set is open and closed. As same as this logic, a whole set $M$ is also open and closed.

Product topology

Let $(M,\mathcal{O}_M)$ and $(N,\mathcal{O}_N)$ be topological spaces. We can equip the product topology on the cartesian product $U:=M\times N$ as follows:

$$ U \in \mathcal{O}_{M\times N} :\Leftrightarrow \forall p=(m,n) \in U: \exists (S,T) \in \mathcal{O}_M \times \mathcal{O}_N : S \times T \subseteq U $$

In this way, we can extend the definition to any finite number of product, e.g., $\mathcal{O}_{\mathrm{std}\ \mathbb{R}^d} := \mathcal{O}_{\mathrm{std}\ \underbrace{\mathbb{R}\times\mathbb{R}\times\cdots\times\mathbb{R}}_{d\text{ times}}}$


Define a sequence as a map $q:\mathbb{N} \rightarrow M$. A sequence $q$ on a topological space $(M,\mathcal{O}_M)$ is said to converge against a limit point $a\in M$ if,

$$ \underbrace{\forall O\in\mathcal{O}_M}_{a\in O\text{, an open neighbourhood of }a}: \exists N\in\mathbb{N}: \forall n>N: q(n) \in O $$

Almost constant sequence

I couldn’t find the agreed terminology of this concept.

  • almost constant
  • eventually constant
  • almost surely constant (probability theory terminology)
  • defnitely constant

I’ll use “almost constant” as same as Frederic said by means of:

$$ \exists N \in \mathbb{N}: \forall n > N : q(n) = a $$


$\displaystyle q(n)=1-\frac{1}{n+1}$ isn

  • not almost constant
  • not convergent in $(\mathbb{R}, \mathcal{P}(\mathbb{R}))$ (because, $\mathcal{P}(\mathbb{R})$ allows to group distant points)
  • convergent in $(\mathbb{R}, \mathcal{O}_{\mathrm{std}\ \mathbb{R})}$


Let $(M,\mathcal{O}_M)$ and $(N,\mathcal{O}_N)$ be topological spaces. A map $\phi:M\rightarrow N$ is called continuous if

$$ \forall V\in\mathcal{O}_N: \phi^{-1}[V] \in \mathcal{O}_M $$

Continuum: “continuity” of real numbers $\mathbb{R}$

Continuity above is defined for a map, not a topological space itself. But still, we have intuition about the “continuity” which the set of real numbers equips. The origin of continuity of real numbers is depend on how you construct the set of real numbers.

Here is my note about continuum.

One way to construct reals is by Cauchy sequences. (We are still able to define “distance/metric” upon $\mathbb{Q}$.)

$\epsilon - \delta$


sepreation properties

$T_1$ (Fréchet), $T_2$ (Hausdorff)

Zariski topology

$T_1$ but not $T_2$


Open cover

also subcover

Heine–Borel theorem

Paracompact space

Connectedness & path-connectedness

Homotopic curves

Fundamental group


Old note

A topological space is an ordered pair $(X,\tau)$, where $X$ is a set and $\tau$ is a collection of subsets of $X$ satisfying the following axioms:

  1. The empty set and $X$ itself belong to $\tau$.:
  2. Any arbitrary (finite or infinite) union of members of $\tau$ belongs to $\tau$.
  3. The intersection of any finite number of members of $\tau$ belongs to $\tau$.

The elements of $\tau$ are called open sets and the collection $\tau$ is called a topology on $X$.

The element of $X$ is usually called a point.