Subspace Topology

Intersected Topology

Let

X

be a topological space with toplogy

𝒯

. If

Y

is a subset of

X

, then we define the toplogy

𝒯

intersected with

Y

as the set

T_{\cap Y} : = {Y \cap U : U \in 𝒯}

Intersected Topology is a Topology

The set

T_{\cap Y}

is a topology on

Y

We'll show that $\emptyset$ and $Y$ are elements of $T_{\cap Y}$ , this can be seen as since $\emptyset \in T$ , then $Y \cap \emptyset = \emptyset \in T_{\cap Y}$ and also since $Y \subseteq X$ , then $Y \cap X = Y$ so $Y \in T_{\cap Y}$

Now we want to prove that $T_{\cap Y}$ is closed under arbitrary unions, which we know because $⋃_{α \in I} (U_{α} \cap Y) = (⋃_{α \in I} U_{α}) \cap Y$ . Similarly we can see that it's closed under finite intersections as $(U_{1} \cap Y) \cap \dots \cap (U_{n} \cap Y) = (U_{1} \cap \dots \cap U_{n}) \cap Y$ .

Therefore we know that $T_{\cap Y}$ is a topology.

Subspace Topology

Since

T_{\cap Y}

forms a topology, we will denote it by

𝒯_{\cap Y}

and call it the subspace topology and that

Y

is a subspace of

X

Subspace Topology by Universal Properties

Given

(X, 𝒯_{X})

and

Y \subseteq X

then the subspace topology is the unique topology such that

The inclusion from $Y$ into $X$ is continuous
Given a function $f : Z \to Y$ , if $ι \circ f$ is continuous, then $f$ is too

We show that the inclusion is continuous, so let $U$ be open in $X$ then consider $ι^{- 1} (U) = Y \cap U$ which is open in the subspace topology so that $ι$ is continuous

Now suppose that $f : Z \to Y$ and that $ι \circ f$ is continuous, let's verify that $f$ is continuous, so let $V$ be open in $Y$ by the very definition this means that there is some $U \in 𝒯_{X}$ such that $V = U \cap Y$ therefore $f^{- 1} (V) = f^{- 1} (U \cap Y) = f^{- 1} (ι^{- 1} (U)) = {(ι \circ f)}^{- 1} (U)$ and therefore is open, as needed.

To see uniquness we suppose that $ℛ_{Y}$ and $𝒮_{Y}$ are both topologies satisfying the conditions, and we'll prove their equal by showing that the identity is a homomorphism. We start by considering the inclusions $ι_{R} : (Y, ℛ_{Y}) \to X$ and $ι_{S} : (Y, 𝒮_{Y}) \to X$ , now suppose our identity is of the form $id : (Y, ℛ_{Y}) \to (Y, 𝒮_{Y})$ then we know that $ι_{S} \circ id = ι_{R}$ , since $ι_{R}$ is continuous, then by the second listed property we deduce that $id$ is continuous, so that $ℛ_{Y} \supseteq 𝒮_{Y}$ , but then by doing the proof again with the roles of the topologies switched we obtain equality.

Basis for the Subspace Topology

Suppose that

ℬ

is a basis for the topology of

X

, then

M : = {Y \cap B : B \in ℬ}

is a basis and it generates

𝒯_{\cap Y}

To prove this statement true, we'll use the handy basis criterion

First note that given any element $Y \cap B \in M$ , we know that it is open with respect to $𝒯_{\cap Y}$ , because $B$ is open with respect to $X$

Now let $Y \cap U$ be an open set of $Y$ which means that $U$ is open in $X$ . Let $p \in Y \cap U$ , since $p \in U$ , $U$ is open in $X$ and $ℬ$ is a basis for $X$ then we know there is some $B \in ℬ$ such that $p \in B \subseteq U$ if that's the case then we know that $p \in Y \cap B \subseteq Y \cap U$ , noting that $Y \cap B \in M$ , therefore we know that $M$ is a basis and it is for $Y$

open in a subspace implies open

Suppose that

Y

is a subspace of

X

and also that

Y

is open in

X

, then any set that's open in

Y

is also open in

X

Suppose that

U

is open in

Y

, thus there is some

V

open in

X

such that

U = Y \cap V

, but since

Y

is open in

X

, then as a finite intersection of open elements, then

U

is also open in

X

, as needed.

TODO

Suppose that

A

is a subspace of

X

and

B

is a subspace of

Y

, then the product topology on

A \times B

is the same as the topology

A \times B

has as a subspace of

X \times Y

We will show that the two topologies are equal by showing that the basis that generate them are equal.

Given the subspace $A$ of $X$ and $B$ of $Y$ , then the product topology of $A \times B$ has a basis of elements of the form $(A \cap U) \times (B \cap V)$ where $U, V$ are open in $X, Y$ respectively.

On the other hand ,an open set in $X \times Y$ is of the form $U \times V$ where $U, V$ are open in $X, Y$ respectively, thus a basis element for $A \times B$ (as a subspace of $X \times Y$ )is a set of the form $(A \times B) \cap (U \times V)$ .

Now to connect the two recall thqt $(A \cap U) \times (B \cap V) = (A \times B) \cap (U \times V)$ . So given a set, it is in the first basis if and only if it is in the second basis, meaning that the two bases are equal, and thus they generate the same topology.

Lines Intersected With Boxes

L

is a straight line in the plane, describe the topology

L

inherits as a subspace of

ℝ_{ℓ} \times ℝ

and as a subspace of

ℝ_{ℓ} \times ℝ_{ℓ}

}

Since we know that a basis for

ℝ_{ℓ} \times ℝ

are the sets of the form

{[a, b) \times (c, d)}

thus we know that the basis for the subspace

L

are those basis elements intersected with

L

, for the moment supposing that

L

is a line which is not vertical, then given a basis element

B

there are a few possibilities for the non-empty intersections:

$L$ hits the left wall of $B$ , then the intersection is a half open interval
$L$ does not hit the left wall of $B$

$L$ hits the top and right, or bottom and right wall, the intersection results in an open interval

Finally note that it's impossible for the line to hit a corner of the square as the open intervals stop that from occuring, therefore the above enumerates all possibilties for a non-vertical line. In other words the basis for this subspace are the open intervals and the half open intervals, since we already know that the standard topology is a subspace of the lower limit topology, this simply generates the lower limit topology, formally we could show that there is a homeomorphism between them by finding the intersection with the x axis, then doing a rotation, but since we are describing, we will not go so far.

When $L$ is vertical, so that $L = {(x_{0}, t) : t \in ℝ}$ it's not hard to see that the map $f ((x_{0}, t)) = t$ and its inverse define a homeomorphism because open sets in $L$ are all just intervals, and $f$ pulls those back to the same interval in $R$ (the other direction is similar) and so the vertical line subspace is the standard topology.

Now if we consider $ℝ_{ℓ} \times ℝ_{ℓ}$ , and look at the line again, if $L$ has positive slope then we get the lower limit topology as we get a basis as the half open intervals, if it is vertical or horizontal, then similarly to the above above, we get the factors, in this case the lower limit topology elsewise the slope is negative and it ends up being the discrete topology, that's because every single point set on $L$ is open, to see why, given a single point $(x, y)$ then it equals $L \cap [x, x + 1) \times [y, y + 1) = {(x, y)}$ thus we have all singletons, so the discrete topology.