Continuity Is a Local Condition
Continuity of a map can be tested one point at a time. At a point \(x\), the test asks whether points sufficiently near \(x\) are sent into any prescribed neighborhood of \(f(x)\). The topology determines what “near” means: in a metric space it is expressed by distance, while in a general topological space it is expressed by open neighborhoods.
Let \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) be topological spaces, let \(f:X\to Y\), and fix \(x\in X\). An open neighborhood of a point is an open set containing that point. In particular, a neighborhood of \(f(x)\) means an open set \(V\in\mathcal{T}_Y\) with \(f(x)\in V\).
The quantifiers matter. The target neighborhood \(V\) is chosen first, and the source neighborhood \(U\) may depend on \(V\). There must be a suitable \(U\) for each open neighborhood \(V\) of \(f(x)\). The condition does not require \(f(U)=V\), or even that \(f(U)\) contain a neighborhood of \(f(x)\); only the inclusion \(f(U)\subseteq V\) is required.
The Preimage Form of the Neighborhood Test
Proof. Suppose first that \(f\) is continuous at \(x\). For an open neighborhood \(V\) of \(f(x)\), choose an open neighborhood \(U\) of \(x\) with \(f(U)\subseteq V\). The inclusion means that every \(u\in U\) satisfies \(f(u)\in V\), so \(u\in f^{-1}(V)\). Therefore \(U\subseteq f^{-1}(V)\), and the preimage contains an open neighborhood of \(x\).
Conversely, suppose that \(f^{-1}(V)\) contains an open neighborhood \(U\) of \(x\) for every open neighborhood \(V\) of \(f(x)\). If \(u\in U\subseteq f^{-1}(V)\), then \(f(u)\in V\). Thus \(f(U)\subseteq V\), which is exactly the neighborhood condition for continuity at \(x\). \(\square\)
This statement does not say that \(f^{-1}(V)\) itself must be open. It says only that the preimage contains an open set around the particular point \(x\). If the condition holds at every \(x\in X\), then each open set \(V\subseteq Y\) has an open preimage: for each \(x\in f^{-1}(V)\), the condition supplies an open neighborhood of \(x\) contained in \(f^{-1}(V)\), and the union of those neighborhoods is \(f^{-1}(V)\). Conversely, if every open set has open preimage, then \(f^{-1}(V)\) is itself a suitable neighborhood of \(x\) whenever \(x\in f^{-1}(V)\). Thus continuity at every point recovers the usual open-preimage definition of a continuous map.
Worked Example: A Metric Estimate for the Squaring Map
Give \(\mathbb{R}\) its standard topology and consider \(f:\mathbb{R}\to\mathbb{R}\), \(f(t)=t^2\). We verify continuity at \(x=2\) using neighborhoods.
Let \(V\) be any open neighborhood of \(f(2)=4\). Since \(V\) is open and contains \(4\), there is some \(\varepsilon>0\) such that \((4-\varepsilon,4+\varepsilon)\subseteq V\). Set \(\delta=\min(1,\varepsilon/5)\), which is positive, and take \(U=(2-\delta,2+\delta)\).
If \(t\in U\), then \(|t-2|<\delta\leq 1\), so \(1<t<3\) and therefore \(|t+2|<5\). Hence
It follows that \(f(t)\in(4-\varepsilon,4+\varepsilon)\subseteq V\) for every \(t\in U\). Thus \(f(U)\subseteq V\), as required. The estimate works for every open neighborhood \(V\) of \(4\), not just for one chosen interval.
Check: \(2\in U\), and \(f(2)=4\in V\). The strict inequality \(|t-2|<\delta\) gives \(|t^2-4|<5\delta\), even when \(\delta=\varepsilon/5\), so the endpoint choice causes no problem.
Metric Spaces and the Epsilon-Delta Test
For metric spaces, the neighborhood condition has a familiar form. Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. A ball of radius \(r>0\) about \(x\) is \(B_{d_X}(x,r)=\{y\in X:d_X(x,y)<r\}\); define balls in \(Y\) in the same way.
Proof. Suppose \(f\) is continuous at \(x\), and let \(\varepsilon>0\). The ball \(B_{d_Y}(f(x),\varepsilon)\) is an open neighborhood of \(f(x)\). By continuity at \(x\), there is an open neighborhood \(U\) of \(x\) such that \(f(U)\subseteq B_{d_Y}(f(x),\varepsilon)\). Because \(U\) is open and contains \(x\), some \(\delta>0\) satisfies \(B_{d_X}(x,\delta)\subseteq U\). If \(d_X(x,y)<\delta\), then \(y\in U\), so \(f(y)\in B_{d_Y}(f(x),\varepsilon)\); equivalently \(d_Y(f(x),f(y))<\varepsilon\).
Conversely, suppose the stated epsilon-delta condition holds, and let \(V\) be any open neighborhood of \(f(x)\). Openness gives an \(\varepsilon>0\) with \(B_{d_Y}(f(x),\varepsilon)\subseteq V\). Choose \(\delta>0\) from the assumed condition and set \(U=B_{d_X}(x,\delta)\). This is an open neighborhood of \(x\). Every \(y\in U\) satisfies \(d_X(x,y)<\delta\), so \(f(y)\in B_{d_Y}(f(x),\varepsilon)\subseteq V\). Thus \(f(U)\subseteq V\), proving continuity at \(x\). \(\square\)
The criterion concerns distances from the fixed points \(x\) and \(f(x)\). It does not require one value of \(\delta\) to work at every point of \(X\); \(\delta\) may depend on both \(x\) and \(\varepsilon\).
Worked Example: A Function Continuous at One Point but Not Another
Give \(\mathbb{R}\) its standard topology and define \(g:\mathbb{R}\to\mathbb{R}\) by \(g(t)=0\) when \(t=0\), and \(g(t)=1\) when \(t\ne0\). We test continuity at \(1\) and at \(0\).
At \(1\), let \(V\) be any open neighborhood of \(g(1)=1\). The interval \(U=(1/2,3/2)\) is an open neighborhood of \(1\), and it does not contain \(0\). Therefore \(g(t)=1\) for every \(t\in U\), so \(g(U)=\{1\}\subseteq V\). The function is continuous at \(1\).
At \(0\), take the open neighborhood \(V=(-1/2,1/2)\) of \(g(0)=0\). Every open neighborhood \(U\) of \(0\) contains an interval \((-r,r)\) for some \(r>0\). The point \(r/2\) belongs to that interval, is nonzero, and hence satisfies \(g(r/2)=1\notin V\). Thus no open neighborhood \(U\) of \(0\) has \(g(U)\subseteq V\), so \(g\) is not continuous at \(0\).
Check: The same target neighborhood \(V=(-1/2,1/2)\) shows the failure at \(0\), while the explicit neighborhood \((1/2,3/2)\) works at \(1\). Continuity at one point therefore does not imply continuity at every point.
A Finite Example Without Metrics
The neighborhood definition does not depend on distances. To see this directly, let \(X=\{u,v\}\) have the indiscrete topology \(\mathcal{T}_X=\{\varnothing,X\}\). Let \(Y=\{a,b\}\) have the Sierpinski topology \(\mathcal{T}_Y=\{\varnothing,\{a\},Y\}\), and define \(f(u)=a\), \(f(v)=b\). These are topological spaces: each listed collection contains the empty set and the whole space, and is closed under arbitrary unions and finite intersections.
Worked Example: Testing Points in Finite Topologies
At \(v\), the only open neighborhood of \(f(v)=b\) in \(Y\) is \(Y\), because the only listed open set containing \(b\) is \(Y\). The only open neighborhood of \(v\) in \(X\) is \(X\). Since \(f(X)=Y\subseteq Y\), the neighborhood condition holds at \(v\).
At \(u\), the set \(\{a\}\) is an open neighborhood of \(f(u)=a\). But the only open neighborhood of \(u\) in \(X\) is \(X\), and \(f(X)=Y\) is not a subset of \(\{a\}\), since \(b\in Y\setminus\{a\}\). No suitable source neighborhood exists, so \(f\) is not continuous at \(u\).
Check: The preimage formulation gives the same result: \(f^{-1}(Y)=X\) contains a neighborhood of \(v\), while \(f^{-1}(\{a\})=\{u\}\) contains no open neighborhood of \(u\) in the indiscrete topology. This example uses only the specified open sets; no metric or distance comparison is involved.
A Common Pitfall: One Neighborhood Is Not Enough
A frequent error is to find one pair \(U,V\) for which \(f(U)\subseteq V\) and conclude that \(f\) is continuous at \(x\). That verifies only one instance of the condition. Continuity requires a suitable source neighborhood for every open neighborhood of the image.
For a concrete wrong argument, use the function \(g\) above at \(0\). One might choose \(V=\mathbb{R}\) and \(U=\mathbb{R}\), observe that \(g(U)\subseteq V\), and claim continuity. The calculation is true but irrelevant to the required test: \(V=\mathbb{R}\) is only one of the open neighborhoods of \(g(0)\). For \(V=(-1/2,1/2)\), every open neighborhood \(U\) of \(0\) contains a nonzero point \(r/2\) for some \(r>0\), and \(g(r/2)=1\notin V\). This particular \(V\) defeats every possible \(U\), proving discontinuity at \(0\).
The local viewpoint is useful because it identifies exactly what must be controlled near a chosen point. In metric spaces, estimates provide the control; in other spaces, the open sets themselves provide it. In either setting, keep the order of the test fixed: choose an arbitrary neighborhood of the image, then find a source neighborhood whose entire image lies inside it.
Check Your Understanding
For each question, identify the relevant open neighborhoods before deciding whether continuity holds.
- State the neighborhood definition of continuity at \(x\), including the order of its quantifiers.
- For \(f:\mathbb{R}\to\mathbb{R}\), \(f(t)=t^2\), use the metric criterion to verify continuity at \(x=0\).
- In the finite-space example, why does the only open neighborhood of \(v\) in \(X\) have to be \(X\)?
- Give a target neighborhood that proves \(g(t)=0\) for \(t=0\) and \(g(t)=1\) otherwise is not continuous at \(0\).
- Explain why finding \(U\) for one open neighborhood \(V\) of \(f(x)\) does not establish continuity at \(x\).