Lecture 1
What is topology as a subject?
Topological spaces
Examples
Lecture 2
Lecture 3
Constructing topological spaces
Subspace topology
Finite products
Lecture 4
Closed sets
We call a subset
$Y\subset X$ closed if $X\backslash Y$ is open
Using this idea there
can be multiple examples which we can see. Like all the elements of a discrete
topology are closed, $\phi , X$ are both closed.
Now let’s have a look on
some of the properties of closed sets.
Theorem:
1. The subsets $\phi$
and $X$ are closed.
2. Finite union of
closed sets is closed.
3. Arbitrary
intersection of closed sets is closed.
Let’s try to prove it,
1. Note that
$X\backslash \phi=X$ and $X\backslash X=\phi$, and both are open
2. Consider the closed
sets $U$ and $V$. Now $X\backslash (U\cup V) = (X\backslash U)\cap(X\backslash
V)$. Now finite intersection of open sets is open.
3. Consider
$\{V_{\alpha}\}$. Now $X\backslash \{\cup V_{\alpha}\}=\cap (X\backslash
V_{\alpha})$. Now arbitrary union of open sets is open.
Definition: We say that $U$ is closed in $Y$ if $U\subseteq
Y$ and $U$ is closed under the subspace topology of $Y$. ($Y\subseteq X$)
Theorem: $U$ is closed in $Y$ iff $\exists$ some closed
$A\subset X$ such that $U=Y\cap A$.
Proof: $\Rightarrow$
We have the subspace
topology as $\tau_{Y}=\{U\cap Y|U\subseteq X\}$. Say $A\subset Y$ is closed.
This means $Y\backslash A\in \tau_{Y}$. Therefore $ Y\backslash A = U\cap Y$.
Now, $Y\backslash (Y\backslash A) = Y\backslash (U\cap Y) = (Y\backslash U) =
(X\backslash U)\cap Y$.
Note that $X\backslash
U$ is closed as $U$ is open.
$\Leftarrow$
Say $U=Y\cap A$ where
$A$ is closed subset of $X$. Now clearly $U\subset Y$. Now we know that $Y$ is
closed in the subspace topology of $Y$. This is because $Y=X\cap Y$ and
$Y\backslash (X\cap Y)=\phi$ which is open as it is present in the topology.
Hence $U$ is closed in the subspace topology of $Y$. Therefore it is a
closed set in $Y$.
Theorem: Suppose $Y\subset X$. If $Y$ is closed in $X$
and $U\subset Y$ is closed in $Y$ then $U$ is closed in $X$
Proof: Say $U$ is closed in $Y$ and $Y$ is closed in
$X$. It means that $U=A\cap Y$ where $A\subset X$ is closed and $X\backslash Y$
is open which means it is in the topology of $X$. We need to show that
$X\backslash U$ is also present in the topology of $X$. Let’s have a look.
$X\backslash (A\cap Y) =
(X\backslash A)\cup (X\backslash Y)$. $X\backslash A$ is open and so is
$X\backslash Y$. Finite union of open sets is open. Therefore $X\backslash U$
is open in $X$. Therefore $U$ is close in $X$.
Now say $U\subset X$ is
closed. Also $U\subset Y$ and $Y\subset X$ is closed. We need to show that $U$
is closed in $Y$. Now $Y\backslash U = (X\backslash U )\cap Y$. Note that
$Y$ is open in $Y$, and $X\backslash U$ is open as well. Therefore $U$ is
closed in $Y$
Closure and interior of
a set
Consider $X$ as a set.
We define closure of a subset $A$ of $X$, $Cl(A)$ as the intersection of all
the closed subsets of $X$ containing $A$. On the other hand, the interior of
$A$, $Int(A)$ is the union of all the subsets of $X$ which are contained in
$A$. From these we get a few properties.
1. If $U\subset Y\subset
X$ and $U$ is open in $Y$, then $U\subset Int(Y)$
2. If $Y\subset V
\subset X$ and $V$ is closed in $X$, then $Cl(Y)\subset V$.
Some Lemmas to keep you
busy:
My bud, try to prove
these lemmas, (Although I will give the proof as well but try it on your own
for a while)
1. $X\backslash
(Cl(A))=Int(X\backslash A)$
2. $X\backslash
(Int(A))=Cl(X\backslash A)$
Proof (Look at these only when you have tried
them already)
1. Now we have seen that
$Cl(A)=\bigcap\limits_{\alpha}K_{\alpha}$ (These sets are closed and contain
$A$). Now $X\backslash \left(\bigcap\limits_{\alpha}K_{\alpha}\right)=
\bigcup\limits_{\alpha}(X\backslash K_{\alpha})$. Note that the sets
$(X\backslash K_{\alpha})$ are open and are the ones that do not contain the
set $A$ which means, they are a subset of $X\backslash A$. Therefore by
definition, it is by definition $Int(X\backslash A)$.
2. For the case of
interior, we have $Y=\bigcup\limits_{\alpha}U_{\alpha}$ and the sets
$U_{\alpha}$ are open in $Y$. Now $X\backslash
\bigcup\limits_{\alpha}U_{\alpha} = \bigcap\limits_{\alpha}(X\backslash
U_{\alpha})$. Now each $ X\backslash U_{\alpha}$ is closed and contains
$X\backslash Y$. Therefore we get $Cl(X\backslash Y)$.
Some more interesting
pieces of stuff about closure property.
We say that a subspace
$Y\subset X$ is dense in $X$ if $Cl(Y)=X$
Fun fact: $\mathbb{Q}$
has no interior but its closure is $\mathbb{R}$.
Say $Y\subseteq X$ and
$A\subseteq Y$. Then $Cl_Y(A)=Cl_X(A)\cap Y$
Proof: We have
$U\subseteq Y$ and $Y\subseteq X$. Now $Cl_Y(X)$ refers to the closure of $U$
in $Y$. Note that $Y\cap Cl_X(A)$ is closed. Now by definition, the closure
must be the smallest closed set containing $A$. Therefore $ Cl_Y(A) \subseteq
Cl_X(A)\cap Y$. Now we have $Cl_Y(X)=C\cap Y$ where $C\subset X$ is a closed
subset. Now $A$ is contained in $Cl_X(A)$ therefore $Cl_X(A)\subset C\cap Y =
U$.
Lecture 7
Metric
topology:
We define a metric $d$ on a set $X$, which satisfies
the following properties
1. $d(x,x)=0$
2.$d(x,y)=d(y,x)$
3. $d(x,z)\leq d(x,y)+d(y,z)$
We note that $x,y\in X$
We won’t be seeing these a lot in the course, but it
is worth having a note.
We call epsilon balls around $x\in X$, the set $B_{\epsilon}(x)=\{y|d(x,y)<\epsilon\}$.
Definition:
The
metric topology defined on a set $X$ is the topology generated the basis $B$
which is simply the collection of all the epsilon balls around $x\in X$.
Definition:
We
call a set metrizable if there exists some metric $d$ which we can be imposed
on the set $X$.
Definition:
We
say that a set $U\subset X$ is open if for every $y\in U$, $\exists \delta>0$
such that $B_{\delta}(x)\subseteq U$. We note that since $\delta$ can be
reduced, restricting $\delta<1$, say, gives us the same metric topology.
Definition:
We
say that a set is bounded if there exists $M$ such that $\forall a_1,a_2\in X$,
we will have $d(a_1,a_2)<M$.
Proposition:
The
metric $\overline{d}=\min\{d(x,y),1\}$ induces the same topology on $X$ as it
is for the case of $d$.
Proof:
Since
for $\delta<1$, we have $d=\overline{d}$, the metric imposed by the two
metrics are the same.
Proposition:
Say
we have two metrics $d$ and $d’$. Say the topologies imposed by them are
respectively are $\tau$ and $\tau ’$. We
say that $\tau’$ is finer than $\tau$ if for each $\epsilon>0$, there exists
$\delta$ such that $B_{\delta,d’}\subseteq B_{\epsilon,d}$.
Proof:
Later XP
Proposition:
The
product topology on $\mathbb{R}$ is the same as the topologies induced by the
Euclidean metric and the box metric.
Proof: It is easy to show that one metric is finer than the other metric using some standard inequalities and the previous lemma. For the second part, we can show that under square topology, if $x\in B_{\epsilon}(x)$, then we can take the set $\prod U_i$ where $U_i=(x_i-\delta,x_i+\delta)$ where $\delta<\epsilon$. Therefore we get a set $U$ such that $x\subset U \subset B_{\epsilon}(x)$. Conversely, for $x\in U$, we can take the ball $B_{\epsilon}(x)=(x_1-\epsilon,x_1+\epsilon)\times\ldots (x_n-\epsilon,x_n+\epsilon)$ where $(x_i-\epsilon,x_i+\epsilon)\subset U_i$.
First countable axiom
Definition:
We
say that at a point $x\in X$, we have a countable basis if there exists
countable collection of neighborhoods around $x$, $\{B_n\}_{n=1}^{\infty}$,
such that for any neighborhood around $x$, $U$, we will have some $B_i\subset U$.
A space is first countable if all the points in it is first countable. Note
that any $B_n$ can be replaced with the intersection of nested neighborhoods.
This will lead us to a few ideas, like
Lemma:
Every metric space is first countable (Just take $\{B_{1/n}\}$).
Lemma:
Say
$\{a_n\}$ converges to $x\in X$ with $a_i\in A$. Then $x\in Cl A$.
Lemma:
Say
$X$ is metrizable and $x\in Cl A$, then there exists a sequence $\{a_n\}$ such
that $a_i\in A$ and converges to $x$.
Lemma:
Suppose
we have a function $f:X\to Y$, then
1. If $f$ is continuous then $f(a_n)$ converges to $f(a)$
if $a_n\to a$
2. If $X$ is first countable and $a_n\to a$ and the
image sequence $f(a_n)\to f(a)$, then $f$ is continuous.
Lecture 8
Recall equivalence relation divides a set into
sections which we call equivalence classes. We define
$[x]=\{y\in
X|x\tilde y\}$
We denote the set of all equivalence classes of $X$
in the following way
$X/\sim
= \{[x]|x\in X\}$
The canonical surjection
looks like $\pi(x)=[x]$
Lemma:
Suppose
we have a surjective function $f:X\to Y$. Let’s define an equivalence relation
$\sim$, such that $x\sim x’$ if and only if $f(x)=f(x’)$. Now there is an
induced bijection given by
$(X/\sim)\to
Y$
Given by $h([x])=f(x)$.
We will have an inverse $h^{-1}(y)$ which will take $y\in Y$ to $f^{-1}(y)$.
That will give us $[x]$ where $x$ can be anything which satisfies $f(x)=y$. The
surjection $f:X\to Y$ factors as a canonical surjection $\pi: X\to (X/\sim)$ and
a bijection $h: (X/\sim)\to Y$.
We will prove this
tomorrow.
Definition:
Suppose
we have a surjective function $f:X\to Y$, we will call the Quotient topology on
$Y$ induced by $X$ as the set of all $V\subset Y$ such that $f^{-1}(V)$ is open
in $X$.
Definition:
We
say that a map $f:X\to Y$, is a Quotient map if a quotient topology can be
imposed on $Y$ from $X$.
Lemma:
Quotient
map is a continuous map.
Lemma:
Consider
a function $f:X\to Y$ where $X$ is a topological space. Then the quotient
topology defined on $Y$ is the finest topology such that $f$ is continuous.
The proof of this lemma
follows from the definition.
Lemma:
A
surjective function $f$ is a quotient map if and only if the following
condition holds: $K\subseteq Y$ is closed in $Y$ if and only if $f^{-1}(K)$ is
closed in $X$.
The proof to this lemma
also follows from the definition.
Definition:
1. We call a function
to be open if $f(U)\in \tau_{Y}$ for $U\in\tau_{X}$
2. $f$ is said to be
closed if $f(C)$ is closed in $Y$ for closed $C\subset X$,
Lemma
1.
Each
surjective open map is Quotient map.
2.
Each
surjective closed map is a Quotient map.
Proofs:
Note:
We
need to show that $V\subseteq Y$ is open iff $f^{-1}(V)$ is open in $X$. Same
is the case for a closed set.
1.
Say
$V\subseteq Y$. If $V$ is open in $Y$, then by continuity, $f^{-1}(V)$ is open
in $X$. Conversely, if $f^{-1}(V)$ is open in $V$, $f(f^{-1}(V))=V$ must be
open in $Y$ as the map is an open map.
2.
Similar
is the case here, just replace open with closed.
Theorem:
Say
we have a quotient map $f:X\to Y$. Say $B\subset Y$ and $A=f^{-1}(B)\subset X$.
Now let’s call the restriction map $g:A\to B$. Then
1. If $A$ is open/closed, then $g$ is a quotient map.
2.If the map $f$ is open/closed then $g$ is a
quotient map.
Theorem:
Say
that we have a quotient map $f:X\to Y$. Say $Z$ is a topological space and $h:X\to
Z$ is a map such that $h(x)=h(x’)$ whenever $f(x)=f(x’)$. Then there exists a
unique map $g:Y\to Z$ such that $h=g\circ f$. Iff $h$ is continuous then $g$ is
continuous. Also, iff it is a quotient map, then $g$ is a quotient map as well.
Proof:
A
few hours later maybe.
Corollary:
If $h:X\to Y$ is a map and $\sim$ is an equivalence relation such that $x\sim y$
iff $h(x)=h(y)$. Say the equivalence classes are $X/\sim$. Then
1. The map $h$ induces a unique bijection $g:X/\sim\to
Y$.
2. $g$ is a homomorphism if $h$ is a quotient map.
3. If $Y$ is Hausdorff, then so is $X/\sim$.
Second
countable
Definition:
We
say that a space $X$ is second countable
if there exists a countable basis which generates $\tau_X$.
Proposition:
If
$X$ is second countable, then it is separable.
.
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