Coordinate Slices and Lines as Subspaces of the Plane

Reading a Plane Neighborhood on a Line

A subset of a line in the plane inherits its open sets from the plane: an open set in the line is the intersection of that line with an open set in \(\mathbb{R}^2\). This means that a neighborhood on the line can be found by taking an ordinary plane neighborhood and keeping only the points that lie on the line. For coordinate axes and other horizontal or vertical lines, Euclidean disks make this operation especially concrete.

Throughout, \(\mathbb{R}^2\) has its Euclidean topology, whose basic open sets are disks \[ B((u,v),r)=\{(x,y)\in\mathbb{R}^2:(x-u)^2+(y-v)^2<r^2\}, \] where \(r>0\). A line \(L\subseteq\mathbb{R}^2\) has the subspace topology: its open sets are the sets \(L\cap U\), where \(U\) is open in \(\mathbb{R}^2\). In particular, the relative neighborhood obtained from a disk is \(L\cap B((u,v),r)\).

The Disk–Line Intersection Formula

Proof. A point on \(H_b\) has the form \((x,b)\). It belongs to the disk exactly when \[ (x-u)^2+(b-v)^2<r^2. \] If \(|v-b|\ge r\), then \((b-v)^2\ge r^2\), so the left side is at least \(r^2\) for every \(x\). The strict disk inequality cannot hold, and the intersection is empty. If \(|v-b|<r\), subtracting \((b-v)^2\) gives \[ (x-u)^2<r^2-(v-b)^2. \] The right side is positive, so taking square roots gives \[ -\sqrt{r^2-(v-b)^2}<x-u<\sqrt{r^2-(v-b)^2}. \] Adding \(u\) gives the stated open interval. For \(V_a\), a point has the form \((a,y)\), and disk membership is \((a-u)^2+(y-v)^2<r^2\). If \(|u-a|\ge r\), this is impossible; otherwise, subtracting \((a-u)^2\), taking square roots and adding \(v\) gives the stated interval in \(y\). \(\square\)

The formula distinguishes three cases. If the distance from the disk center to the line is less than \(r\), the intersection is an open interval on the line. If that distance equals \(r\), the disk is tangent to the line at a point on its boundary, but the open disk does not contain that point, so the intersection is empty. If the distance is greater than \(r\), the disk misses the line altogether.

Relative Neighborhoods Are Open Intervals

The formula gives the shape of a disk intersection, but a disk centered at a point of the line gives a particularly simple neighborhood. For \(p=(t,b)\in H_b\) and \(r>0\), the horizontal offset is zero, so \[ B((t,b),r)\cap H_b=\{(x,b):t-r<x<t+r\}. \] Likewise, for \(p=(a,t)\in V_a\), \[ B((a,t),r)\cap V_a=\{(a,y):t-r<y<t+r\}. \] These are open intervals along their respective lines, centered at the chosen point.

Proof. First suppose \(A\) is open in \(L\), and take \(p\in A\). By the definition of the subspace topology, \(A=L\cap U\) for some open set \(U\subseteq\mathbb{R}^2\). Since \(p\in U\) and plane open disks form a basis, there is an \(r>0\) such that \(p\in B(p,r)\subseteq U\). Therefore \(L\cap B(p,r)\subseteq A\), and the centered-disk formulas above show that this is an open interval on \(L\) containing \(p\). Conversely, suppose every \(p\in A\) is contained in an open interval on \(L\) that is itself contained in \(A\). Each such interval is \(L\cap B(p,r)\) for an appropriate \(r>0\), so it is open in \(L\). Their union is \(A\); a union of open sets in a topology is open. If \(A\) is empty, it is open by the topology axioms and the pointwise condition is vacuous. This proves both directions. \(\square\)

There is also a direct coordinate description. On \(H_b\), the coordinate \(x\) identifies points \((x,b)\) with real numbers; on \(V_a\), the coordinate \(y\) identifies \((a,y)\) with real numbers. Under either identification, relative open intervals on the line become ordinary open intervals in \(\mathbb{R}\). The proposition shows that a subset of the line is relatively open precisely when its coordinate values form an open subset of \(\mathbb{R}\). This conclusion follows from the interval neighborhood test, not from treating the line as open in the plane: a line itself is not an open subset of \(\mathbb{R}^2\).

Common Pitfall: Keeping the Full Radius

A disk centered on the line cuts out an interval with half-width \(r\). But if the center is off the line, some of the radius is already used by the perpendicular distance from the center to the line. The remaining half-width is \(\sqrt{r^2-d^2}\), where \(d\) is that perpendicular distance, not \(r\).

For example, a tempting but incorrect calculation for \(B((0,3),5)\cap H_0\) is to write \(\{(x,0):-5<x<5\}\), using the full radius as the horizontal half-width. Substitution into the disk condition gives \(x^2+9<25\), or \(x^2<16\), which is equivalent to \(-4<x<4\). Thus the correct intersection is \(\{(x,0):-4<x<4\}\). The proposed endpoint \(x=5\) has squared distance \(5^2+3^2=34>25\), so it is not even on the disk boundary; this directly exposes the error.

When using the formula, first compare the perpendicular offset with the radius. Only if the offset is strictly smaller should you calculate a nonempty interval. Equality means an empty intersection for an open disk, not a singleton. Finally, use the coordinate of the disk center along the line as the interval’s center: \(u\) on a horizontal line and \(v\) on a vertical line.