Building the Standard Topology from Intervals
The standard topology on the real line is built from open intervals. Its open sets include intervals such as \((2,5)\), but they are not limited to single intervals: unions of open intervals can have several separated pieces, or can cover all of \(\mathbb{R}\). The defining feature is that the set can be written as a union of intervals whose endpoints are excluded.
We use the notation \((a,b)=\{x\in\mathbb{R}:a<x<b\}\), where \(a,b\in\mathbb{R}\) and \(a<b\). These intervals, and all unions of them, are subsets of \(\mathbb{R}\). In particular, the empty union is \(\varnothing\). A topology on a set must contain the empty set and the whole set, be closed under arbitrary unions, and be closed under finite intersections.
This definition does not require a set to be one interval. Nor does it require a union to be finite: an open set may be a union of infinitely many intervals. We first verify that this definition really gives a topology.
Proof. The empty set is the empty union of open intervals, so \(\varnothing\in\mathcal{T}_{\mathrm{std}}\). Also, for each \(n\in\mathbb{N}\), \((-n,n)\) is an open interval, and \[ \mathbb{R}=\bigcup_{n\in\mathbb{N}}(-n,n). \] Indeed, every real number \(x\) belongs to \((-n,n)\) for any integer \(n>|x|\), and every interval in this union is a subset of \(\mathbb{R}\). Thus \(\mathbb{R}\in\mathcal{T}_{\mathrm{std}}\).
For closure under arbitrary unions, let \(\{U_\alpha:\alpha\in J\}\) be any collection of sets in \(\mathcal{T}_{\mathrm{std}}\), where \(J\) is an index set. Each \(U_\alpha\) is a union of open intervals. Taking the union over all \(\alpha\) simply collects all the intervals appearing in all these representations. Therefore \(\bigcup_{\alpha\in J}U_\alpha\) is again a union of open intervals and belongs to \(\mathcal{T}_{\mathrm{std}}\). This also covers the empty collection, whose union is \(\varnothing\).
For closure under intersections, first consider two open intervals \((a,b)\) and \((c,d)\). Their intersection is empty if \(\max(a,c)\geq\min(b,d)\). If \(\max(a,c)<\min(b,d)\), then \[ (a,b)\cap(c,d)=(\max(a,c),\min(b,d)), \] which is an open interval. Now let \(U=\bigcup_{i\in I}I_i\) and \(V=\bigcup_{j\in J}J_j\), with each \(I_i,J_j\) an open interval. Membership in both \(U\) and \(V\) means membership in at least one \(I_i\) and at least one \(J_j\), so \[ U\cap V=\bigcup_{(i,j)\in I\times J}(I_i\cap J_j). \] Each intersection on the right is either empty or an open interval. Hence \(U\cap V\in\mathcal{T}_{\mathrm{std}}\). Repeating this argument proves closure under any positive finite number of intersections; the intersection of zero sets is \(\mathbb{R}\), already shown to belong to the collection. All topology axioms hold. \(\square\)
Unions Can Have Different Shapes
A union of intervals need not itself be an interval. It may have gaps, and the intervals in its representation may overlap or be disjoint. The topology is defined by the existence of a representation as a union, not by a requirement that the resulting set look like one uninterrupted interval.
Worked Example: Two Separated Open Pieces
Consider \(U=(-3,-1)\cup(2,5)\). Each of \((-3,-1)\) and \((2,5)\) is an open interval, so their union belongs to \(\mathcal{T}_{\mathrm{std}}\). The gap \([-1,2]\) is not part of \(U\): the endpoints \(-1\) and \(2\) are excluded from their respective intervals, and every point strictly between them is in neither interval.
For example, \(-2\in(-3,-1)\) because \(-3<-2<-1\); \(4\in(2,5)\) because \(2<4<5\); and \(0\notin U\) because \(-2<0<2\), so \(0\) lies in neither interval. The set has two separated pieces, but it is open because it is a union of open intervals.
Check: the displayed representation uses exactly the two intervals \((-3,-1)\) and \((2,5)\). Their union contains every point of either interval and no point outside them, so it is precisely \(U\).
Worked Example: The Real Line and a Punctured Line
The whole real line is open in the standard topology, since \[ \mathbb{R}=\bigcup_{n\in\mathbb{N}}(-n,n). \] For each \(x\in\mathbb{R}\), the Archimedean property gives an integer \(n>|x|\), so \(-n<x<n\) and \(x\) is in this union. Conversely, every \((-n,n)\) is contained in \(\mathbb{R}\).
The set \(\mathbb{R}\setminus\{0\}\) is also open. An explicit representation is \[ \mathbb{R}\setminus\{0\} =\left(\bigcup_{n\in\mathbb{N}}(-n,0)\right) \cup \left(\bigcup_{n\in\mathbb{N}}(0,n)\right). \] Every interval on the right excludes \(0\), so the union contains no \(0\). If \(x<0\), choose \(n\in\mathbb{N}\) with \(n>|x|\); then \(-n<x<0\). If \(x>0\), choose \(n\in\mathbb{N}\) with \(n>x\); then \(0<x<n\). Thus every nonzero real number belongs to the union.
Check: the right-hand side contains exactly the negative real numbers and the positive real numbers, and contains neither \(0\). It is therefore exactly \(\mathbb{R}\setminus\{0\}\), not all of \(\mathbb{R}\).
Sets That Are Not Open
To show that a set is not standard-open, it is enough to show that no union of open intervals can equal it. If a point belongs to such a union, it must belong to at least one interval used in the union. That interval also contains points on both sides of the point, because its endpoints are strictly to either side. This observation is useful when a set includes an endpoint but excludes nearby points beyond it.
Worked Example: An Included Endpoint
Consider \(A=(-2,1]\). We show that \(A\) is not open in the standard topology. Suppose, for contradiction, that \(A\) were a union of open intervals. Since \(1\in A\), at least one interval \((a,b)\) in that union would contain \(1\). Containment means \(a<1<b\). The number \(y=(1+b)/2\) then satisfies \(1<y<b\), so \(y\in(a,b)\) and hence \(y\in A\). But \(A=(-2,1]\) contains no number greater than \(1\). This contradiction proves that \(A\) is not open.
The issue is specifically the included endpoint \(1\). The points of \((-2,1)\) lie inside an open interval contained in \(A\), but any open interval containing \(1\) also contains points greater than \(1\), which are not in \(A\).
Check: the value \(y=(1+b)/2\) is strictly greater than \(1\), since \(b>1\); it is strictly less than \(b\), since \(1<b\). It therefore lies in the interval chosen to contain \(1\), while it does not lie in \((-2,1]\).
Worked Example: The Integers
The set \(\mathbb{Z}\) is not open in the standard topology. Suppose it were a union of open intervals. Because \(0\in\mathbb{Z}\), some interval \((a,b)\) in that union would contain \(0\), so \(a<0<b\). Define \(t=\min(b,1)/2\). Since \(b>0\), we have \(t>0\). If \(b\leq1\), then \(t=b/2<b\); if \(b>1\), then \(t=1/2<b\). Thus \(0<t<b\), and \(a<0<t\) gives \(t\in(a,b)\). Also \(t\leq1/2<1\), so \(t\) is not an integer. The interval in the proposed union would therefore include a point not in \(\mathbb{Z}\), a contradiction.
In particular, no singleton \(\{m\}\), with \(m\in\mathbb{Z}\), is open in the standard topology: if it were open, translating the same interval argument from \(0\) to \(m\) would force it to contain points other than \(m\).
Check: in the case \(b\leq1\), \(t=b/2\) is positive and less than \(b\), and in the case \(b>1\), \(t=1/2\) is positive and less than \(b\). In both cases \(t<1\), so \(t\notin\mathbb{Z}\), as required for the contradiction.
A Comparison with the Discrete Topology
On the same underlying set \(\mathbb{R}\), the discrete topology declares every subset open. That differs from the standard topology: some sets are open in both, while some are open only in the discrete topology. A set being open in one topology does not automatically make it open in another topology on the same set.
Worked Example: A Singleton and an Interval
Compare \((2,4)\) and \(\{2\}\). The set \((2,4)\) is open in the standard topology because it is itself an open interval. It is also open in the discrete topology because every subset of \(\mathbb{R}\) is open there.
The set \(\{2\}\) is open in the discrete topology, again because every subset is open. It is not open in the standard topology. Indeed, if it were a union of open intervals, one of those intervals would have to contain \(2\). Any interval \((a,b)\) containing \(2\) has \(a<2<b\), and \(s=(2+b)/2\) satisfies \(2<s<b\). Thus \(s\in(a,b)\) but \(s\ne2\), so that interval cannot be contained in \(\{2\}\). No union of such intervals can equal \(\{2\}\).
Conclusion: \(\{2\}\) is open in the discrete topology but not in the standard topology; \((2,4)\) is open in both. The comparison depends on checking each set against the definition of openness in each topology.
Check: because \(b>2\), the number \(s=(2+b)/2\) satisfies \(s>2\), and because \(2<b\), it satisfies \(s<b\). Hence it is a point of \((a,b)\) distinct from \(2\), verifying the claimed contradiction.
A Common Mistake: Treating Excluded Endpoints as Included
When combining intervals, do not silently add a missing endpoint. For example, it is incorrect to calculate \((0,1)\cup(1,2)=(0,2)\). The point \(1\) belongs to \((0,2)\), but it belongs to neither \((0,1)\) nor \((1,2)\). The correct equality is \[ (0,1)\cup(1,2)=(0,2)\setminus\{1\}. \] This union is still open in the standard topology: it is already expressed as a union of two open intervals. Being open does not mean being a single interval.
The same endpoint care distinguishes \((0,1)\) from \((0,1]\). The first is an open interval and is open in the standard topology. The second includes \(1\) while excluding every point greater than \(1\), and the included-endpoint argument above shows that it is not open.
These examples illustrate the definition in both directions: a displayed union of open intervals certifies openness, while a proposed included point can rule out openness if every interval containing it would also bring in points outside the set.
Check Your Understanding
For each question, use the definition in terms of unions of open intervals. Keep track of whether each endpoint is included.
- Is \((-4,-2)\cup(3,6)\) open in the standard topology? Give a representation that verifies your answer.
- Is \([-1,2)\) open in the standard topology? Identify the point that decides the answer and justify it.
- Write \(\mathbb{R}\setminus\{3\}\) as a union of open intervals.
- Which of \((5,7)\) and \(\{5\}\) is open in the discrete topology but not in the standard topology?
- Explain why \((1,2)\cup(2,3)\) is not equal to \((1,3)\), and state whether the union is open.