# 2.03 Subspace topology

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.