# 2.03 Subspace topology

## Video

Below the video you will find accompanying notes and some pre-class questions.

- Previous video:
**Bases, metric and subspace topologies**. - Next video:
**Connectedness, path-connectedness**. - Index of all lectures.

## Notes

### Subspace topology

*(0.00)* Let \(X\) be a topological space (write \(T\) for the
topology on \(X\)). Suppose that \(Y\subset X\) is a subset. Define
the *subspace topology* on \(Y\) by declaring a subset \(U\subset
Y\) to be open in the subspace topology if and only if there exists
an open subset \(V\subset X\) such that \(U=V\cap Y\).

*(1.37)* Let \(X\) be the plane \(\mathbf{R}^2\) equipped with the
metric topology and let \(Y\) be a curve in \(X\). Take an open ball
in \(X\) and intersect it with \(Y\): you might get something like
an open interval in the curve. This is then open in the subspace
topology on \(Y\).

*(2.34)* Let \(X\) be the plane \(\mathbf{R}^2\) again and let
\(Y=[0,1]\times[0,1]\) be the *closed* square in \(X\). In the
subspace topology on \(Y\), the subset \(Y=[0,1]\times[0,1]\subset
Y\) is *open* (**even though it's closed as a subset of \(X\)**). Of
course, this must be true: otherwise the subspace topology would not
satisfy the axioms of a topology. To see in this case why \(Y\) is
open, take an open ball \(V=B_0(2)\) of radius 2 centred at \(0\) in
\(X\): we have \(Y=V\cap Y\), so \(Y\) is open in the subspace
topology.

*(4.06)* The subspace topology satisfies the axioms for a topology.

Exercise.

### More examples

We can now see many of our favourite mathematical objects as topological spaces.

*(4.28)* The circle \(S^1=\{z\in\mathbf{C}\ :\ |z|=1\}\) is a subset
of the complex plane so it inherits a subspace topology.

*(5.20)* The \(n\)-sphere is the subset
\(\{(x_0,\ldots,x_n)\in\mathbf{R}^{n+1}\ :\ \sum_{k=0}^nx_k^2=1\}\)
inherits a subspace topology from \(\mathbf{R}^{n+1}\)

*(6.07)* The torus \(T^2\) is a subset of \(\mathbf{R}^3\) so it
inherits a topology. Note that I am only talking about the *surface*
of the torus (with longitude and latitude coordinates) so this is a
2-dimensional space. More generally, any closed orientable surface
(genus 2, genus 3,…) can be embedded in \(\mathbf{R}^3\) so
inherits a topology.

*(7.08)* The torus can also be embedded in \(\mathbf{R}^4\). It
inherits another topology from this embedding, but this topology is
*homeomorphic* to the topology it gets from \(\mathbf{R}^3\)
(isomorphic in the category of topological spaces). To see how the
torus embeds in \(\mathbf{R}^4\), let \((\theta,\phi)\) be
longitude/latitude coordinates on the torus and embed this point as
\((\cos\theta,\sin\theta,\cos\phi,\sin\phi)\in\mathbf{R}^4\). Although
the torus in \(\mathbf{R}^3\) and the torus in \(\mathbf{R}^4\) are
homeomorphic (the same topological space) the *geometries* they
inherit from their ambient spaces are very different (in
\(\mathbf{R}^3\) the torus has *Gaussian curvature* which varies
from point to point (from positive to negative); in \(\mathbf{R}^4\)
the Gaussian curvature of the torus is zero).

### Properties of the subspace topology

*(9.45)* Let \(X\) be a topological space with topology \(T\) and
\(Y\subset X\) be a subset. Write \(S\) for the subspace topology on
\(Y\). Let \(i\colon Y\to X\) be the inclusion map. Then:

- \(i\) is continuous with respect to the topology \(S\) on \(Y\) and \(T\) on \(X\);
- moreover, \(S\) is the
*coarsest*topology on \(Y\) for which \(i\) is continuous. That is, if \(S'\) is another topology on \(Y\) such that \(i\colon(Y,S')\to (X,T)\) is continuous, then \(S\subset S'\).

*(12.21)*Let \(V\subset T\) be an open set and let \(i^{-1}(V)\) be its preimage. We have \(i^{-1}(V)=Y\cap V\), which tells us that \(i^{-1}(V)\) is open in the subspace topology on \(Y\). Therefore \(i\) is continuous.*(13.53)*If \(i\) is continuous for \(S'\) then for any \(V\subset X\), \(i^{-1}(V)=V\cap Y\subset Y\) is open in \(S'\). But this means that \(S\subset S'\).

*(15.08)* One can often define a topology on a space \(Y\) by giving a
map \(f\colon Y\to X\) and asking for the coarsest topology on \(Y\)
making \(f\) continuous, or giving a map \(f\colon X\to Y\) and asking
for the finest topology on \(Y\) making \(f\) continuous.

The product topology on \(Y_1\times Y_2\) is the coarsest topology on \(Y_1\times Y_2\) making both projection maps \(p_k\colon Y_1\times Y_2\to Y_k\), \(p_k(y_1,y_2)=y_k\), continuous.

*(16.36)* The maps \(\sin\theta\colon S^1\to\mathbf{R}\) and
\(\cos\theta\colon S^1\to\mathbf{R}\) are continuous maps.

The inclusion map \(i\colon S^1\to\mathbf{R}^2\) is continuous. The projections \(p_k\colon\mathbf{R}^2\to\mathbf{R}\) onto the \(x_1\) and \(x_2\) axes are continuous. Since \(\cos\theta=p_1\circ i\) and \(\sin\theta=p_2\circ i\) are compositions of continuous maps, they are continuous.

Looking back at the embedding of the torus into \(\mathbf{R}^4\), we see that we are thinking of the torus as \(S^1\times S^1\) (coordinates \((\theta,\phi)\)) and the inclusion map we write is continuous with respect to this product topology. This means that the product topology contains the subspace topology (by the lemma above). In fact, when we talk more about homeomorphisms, we will see that the product topology on \(S^1\times S^1\) is homeomorphic to the subspace topology it inherits from \(\mathbf{R}^4\).

## Pre-class questions

- Show that the subspace topology satisfies the axioms for a topology.

## Navigation

- Previous video:
**Bases, metric and subspace topologies**. - Next video:
**Connectedness, path-connectedness**. - Index of all lectures.