Three Questions That Must Be Kept Separate
When a set is described as open, ask: open in which space? When a function is called continuous, ask: which topologies are on its domain and codomain? And when a continuous function is bijective, do not assume its inverse is continuous. These questions are related, but each requires its own check.
For a subset \(A\) of a topological space \((X,\mathcal{T})\), the subspace topology on \(A\) consists of the sets \(A\cap U\), where \(U\in\mathcal{T}\). A set open in this topology is called relatively open in \(A\). It need not be open in \(X\). To discuss continuity, we use the neighborhood definition: \(f:X\to Y\) is continuous at \(x\in X\) if, for every open neighborhood \(V\) of \(f(x)\) in \(Y\), there is an open neighborhood \(U\) of \(x\) in \(X\) such that \(f(U)\subseteq V\). It is continuous if it is continuous at every point.
Relative Openness Is Not Ambient Openness
Proof. The first statement is the definition of the subspace topology. If \(U\) is open in \(X\), then \(A\cap U\) is one of the sets defining that topology, so it is relatively open in \(A\). This does not show that \(A\cap U\) is open in \(X\), because \(A\) itself need not be open in \(X\). \(\square\)
Worked Example: An Interval and a False Relative-Openness Claim
Let \(A=[0,2]\) with the subspace topology inherited from \(\mathbb{R}\). Decide whether \([1,2]\) and \((1,2]\) are relatively open in \(A\), and whether \((1,2]\) is open in \(\mathbb{R}\).
For \((1,2]\), take the ambient open set \(U=(1,3)\). Then \(A\cap U=[0,2]\cap(1,3)=(1,2]\). Thus \((1,2]\) is relatively open in \(A\). It is not open in \(\mathbb{R}\): it contains \(2\), but every open interval around \(2\) contains points greater than \(2\), which are not in \((1,2]\).
Now test \([1,2]\). If it were relatively open, there would be an ambient open set \(U\) containing \(1\) such that \(A\cap U\subseteq[1,2]\). Since \(U\) is open and contains \(1\), it contains an interval \((1-\delta,1+\delta)\) for some \(\delta>0\). Set \(t=\min(\delta/2,1/2)\). Then \(0<t<\delta\), and \(1-t\in A\cap U\), but \(1-t<1\), so \(1-t\notin[1,2]\). This contradicts \(A\cap U\subseteq[1,2]\). Hence \([1,2]\) is not relatively open in \(A\).
Check: The endpoint \(2\) does not prevent relative openness of \((1,2]\), because a sufficiently small ambient neighborhood of \(2\), after intersection with \(A\), stays at or below \(2\). At \(1\), however, every ambient neighborhood meets \(A\) on its left, which is why \([1,2]\) fails the relative-open test there.
A useful special case follows directly from the proposition: if \(A\) is itself open in \(X\), then every relatively open subset \(W=A\cap U\) is ambient open, since it is the intersection of two ambient open sets. Without that extra hypothesis, relative openness alone says nothing of the kind.
Continuity Depends on the Codomain Topology
The formula for a function does not determine whether it is continuous. The neighborhood test examines open neighborhoods in the codomain, so changing the codomain topology can change the answer even if the underlying sets and the function remain fixed.
Worked Example: One Identity Function, Two Codomain Topologies
Let \(X=\{a,b\}\) have the indiscrete topology \(\{\varnothing,X\}\). On the same underlying set \(Y=\{a,b\}\), compare two choices of codomain topology: the indiscrete topology and the discrete topology \(\mathcal{P}(Y)\). Consider the identity assignment \(f(a)=a\), \(f(b)=b\).
With the indiscrete topology on \(Y\), the only open neighborhood of either image point is \(Y\). For each \(x\in X\), choose \(U=X\), which is an open neighborhood of \(x\); then \(f(U)=Y\subseteq Y\). Thus \(f\) is continuous for this codomain topology.
Now give \(Y\) the discrete topology. The set \(\{a\}\) is an open neighborhood of \(f(a)=a\). The only open neighborhood of \(a\) in the indiscrete domain \(X\) is \(X\), and \(f(X)=\{a,b\}\) is not contained in \(\{a\}\). The neighborhood condition fails at \(a\), so \(f\) is not continuous for this codomain topology.
Check: The function has not changed; only the topology on its codomain has changed. The two continuity conclusions differ because the discrete codomain has a smaller neighborhood \(\{a\}\) that the indiscrete domain cannot map locally into.
In particular, a statement such as “this formula is continuous” is incomplete unless the topologies on both spaces are understood. For a map into a subspace, the codomain neighborhoods are relative neighborhoods, not automatically the ambient open sets themselves. The neighborhood test must use the topology of the stated codomain.
A Continuous Bijection Need Not Be a Homeomorphism
Bijectivity guarantees that an inverse function exists as a set map. It gives no continuity guarantee for that inverse. The following criterion isolates the additional condition needed. A map \(f:X\to Y\) is called open if \(f(U)\) is open in \(Y\) for every open \(U\subseteq X\).
Proof. Suppose first that \(f\) is open. Let \(y\in Y\), and let \(U\) be any open neighborhood of \(f^{-1}(y)\) in \(X\). Because \(f\) is open, \(f(U)\) is open in \(Y\); because \(f^{-1}(y)\in U\), we have \(y\in f(U)\). Also, bijectivity gives \(f^{-1}(f(U))=U\). Thus the open neighborhood \(f(U)\) of \(y\) maps under \(f^{-1}\) into \(U\). This is the neighborhood condition for \(f^{-1}\) to be continuous at \(y\). Since \(y\) was arbitrary, \(f^{-1}\) is continuous.
Conversely, suppose \(f^{-1}\) is continuous, and take an open set \(U\subseteq X\). For each \(y\in f(U)\), bijectivity gives a unique \(x\in U\) with \(f(x)=y\). Continuity of \(f^{-1}\) at \(y\), applied to the open neighborhood \(U\) of \(f^{-1}(y)=x\), gives an open neighborhood \(V_y\) of \(y\) such that \(f^{-1}(V_y)\subseteq U\). If \(z\in V_y\), then \(f^{-1}(z)\in U\), so \(z=f(f^{-1}(z))\in f(U)\). Therefore \(V_y\subseteq f(U)\). Every point \(y\in f(U)\) has an open neighborhood contained in \(f(U)\), so \(f(U)\) is open. Hence \(f\) is open. \(\square\)
Worked Example: A Continuous Bijection with a Discontinuous Inverse
Let the domain \(X=\{a,b\}\) have the discrete topology and the codomain \(Y=\{a,b\}\) have the indiscrete topology. Let \(f:X\to Y\) be the identity function. Verify continuity, bijectivity, and failure of inverse continuity.
The function is a bijection: each of \(a\) and \(b\) has exactly one preimage, itself. To check continuity using neighborhoods, fix \(x\in X\). The only open neighborhood of \(f(x)\) in the indiscrete space \(Y\) is \(Y\). The set \(X\) is an open neighborhood of \(x\), and \(f(X)=Y\subseteq Y\). Thus \(f\) is continuous at both points, hence continuous.
The inverse is again the identity assignment, now from the indiscrete space \(Y\) to the discrete space \(X\). At \(a\), the set \(\{a\}\) is an open neighborhood of \(f^{-1}(a)\) in \(X\). The only open neighborhood of \(a\) in \(Y\) is \(Y\), and \(f^{-1}(Y)=\{a,b\}\) is not contained in \(\{a\}\). The inverse is not continuous at \(a\).
The theorem gives a second check: the open set \(\{a\}\) in \(X\) has image \(f(\{a\})=\{a\}\), which is not open in the indiscrete space \(Y\). Thus \(f\) is not open, consistent with the discontinuity of its inverse.
Common Mistakes and a Reliable Checklist
Common mistake: replacing “relatively open” by “open in the ambient space.” A mistaken calculation might say, “\([1,2]\) is open in \(A=[0,2]\) because it includes the endpoint \(2\), where the subspace stops.” That argument overlooks the other endpoint: every relative neighborhood of \(1\) contains points of \(A\) less than \(1\). The correction is to test each point using \(A\cap U\) for ambient open neighborhoods \(U\). In this example, \((1,2]=A\cap(1,3)\) is relatively open, while \([1,2]\) is not.
Common mistake: checking continuity without naming the codomain topology. As the identity example showed, the same function from the same domain can be continuous for one codomain topology and fail for another. Write down the open neighborhood of the image point that must be tested, and then find an open neighborhood of the original point whose image lies inside it. If no such domain neighborhood exists, continuity fails at that point.
Common mistake: treating a continuous bijection as automatically invertible continuously. Bijectivity gives an inverse as a function, not a neighborhood condition for that function. To prove a bijection is a homeomorphism, check continuity of the inverse separately, or use the theorem: verify that the original bijection maps every open set to an open set. For a claimed counterexample, exhibit one open set whose image is not open, as in the finite example.
Record the space in which the set is being tested. If it is a subset \(W\subseteq A\), seek an ambient open \(U\) with \(W=A\cap U\).
State both topologies. At each point under consideration, start with an open neighborhood in the codomain and find a suitable open neighborhood in the domain.
Do not stop at continuity and bijectivity. Check inverse continuity, or prove that images of open domain sets are open in the codomain.
Check Your Understanding
For each question, identify the space and topology used in every openness or continuity claim.
- In \(A=[0,2]\subseteq\mathbb{R}\), is \([0,1)\) relatively open? Is it open in \(\mathbb{R}\)? Justify both answers.
- Give an ambient open set \(U\subseteq\mathbb{R}\) such that \(A\cap U=(0,2]\) for \(A=[0,2]\), or explain why none exists.
- For the identity map from a two-point indiscrete space to the same set with the discrete topology, identify an image point and codomain neighborhood witnessing failure of continuity.
- State the additional condition in the theorem that makes a continuous bijection have a continuous inverse, and prove why that condition suffices.
- In the finite continuous-bijection example, identify an open set whose image is not open and use it to verify that the map is not open.