Two Definitions, One Notion
For maps between metric spaces, continuity can be described locally, by controlling the distance between input and output, or globally, by asking how the map acts on open sets. The two descriptions appear different: one refers to points and positive distances, while the other refers to whole subsets. We prove that they are equivalent when the local condition holds at every point of the domain.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. We use \(B_{d_X}(x,r)=\{z\in X:d_X(x,z)<r\}\) for an open ball in \(X\), and use the corresponding notation for \(Y\). The metric topology consists of the sets \(U\) such that for every \(x\in U\), some open ball centered at \(x\) is contained in \(U\). In particular, metric balls are open: if \(z\in B_{d_X}(x,r)\), then \(s=r-d_X(x,z)>0\), and the triangle inequality gives \(B_{d_X}(z,s)\subseteq B_{d_X}(x,r)\).
The quantifiers matter. In pointwise continuity, \(\delta\) may depend on both \(x\) and \(\varepsilon\); it need not work for all points of \(X\). Open-set continuity makes one statement about every open subset of the target. The equivalence below explains exactly how these two conditions translate into each other.
The Equivalence Theorem
Proof. First suppose \(f\) is epsilon-delta continuous at every point. Let \(V\subseteq Y\) be open. If \(f^{-1}(V)=\varnothing\), then it is open. Otherwise, take any \(x\in f^{-1}(V)\), so \(f(x)\in V\). Since \(V\) is open, there is an \(\varepsilon>0\) such that \(B_{d_Y}(f(x),\varepsilon)\subseteq V\). Continuity at \(x\) gives a \(\delta>0\) such that \(d_X(x,z)<\delta\) implies \(d_Y(f(x),f(z))<\varepsilon\). Thus every \(z\in B_{d_X}(x,\delta)\) satisfies \(f(z)\in V\), so \(B_{d_X}(x,\delta)\subseteq f^{-1}(V)\). We have found such a ball around every point of \(f^{-1}(V)\); by the definition of the metric topology, \(f^{-1}(V)\) is open.
Conversely, suppose the preimage of every open subset of \(Y\) is open in \(X\). Fix \(x\in X\) and \(\varepsilon>0\). The ball \(B_{d_Y}(f(x),\varepsilon)\) is open in \(Y\), so its preimage is open in \(X\). That preimage contains \(x\), since \(f(x)\in B_{d_Y}(f(x),\varepsilon)\). By openness, there is a \(\delta>0\) such that \(B_{d_X}(x,\delta)\subseteq f^{-1}(B_{d_Y}(f(x),\varepsilon))\). Therefore, whenever \(d_X(x,z)<\delta\), we have \(d_Y(f(x),f(z))<\varepsilon\). This proves continuity at \(x\), and \(x\) was arbitrary. If \(X\) is empty, the pointwise condition is vacuous and every preimage is empty, so the equivalence holds in that case as well. \(\square\)
The forward proof uses the openness of the target set to find a ball around the output, then uses continuity to pull that control back to a ball around the input. The reverse proof needs only target balls: openness of their preimages supplies exactly the input neighborhoods required by the epsilon-delta definition.
Proof. If \(f\) is continuous, the theorem applies to every open ball, since open balls are open sets. Conversely, suppose the preimage of every open ball in \(Y\) is open. Let \(V\subseteq Y\) be open. For each \(y\in V\), there is a radius \(r_y>0\) such that \(B_{d_Y}(y,r_y)\subseteq V\). Then \(f^{-1}(V)=\bigcup_{y\in V} f^{-1}(B_{d_Y}(y,r_y))\): membership in the left side means \(f(x)\) is some \(y\in V\), and hence \(x\) belongs to the corresponding preimage on the right; membership in any set on the right implies \(f(x)\in V\). Each set in this union is open by the assumption, and an arbitrary union of open sets is open. Thus every open \(V\) has open preimage, and the theorem gives continuity. If \(V=\varnothing\), the union is empty and is open as well. \(\square\)
Worked Examples
Worked Example: The Identity Map
Let \((X,d)\) be any metric space and let \(\operatorname{id}_X:X\to X\) be the identity map, defined by \(\operatorname{id}_X(x)=x\). Verify both continuity descriptions.
Fix \(x\in X\) and \(\varepsilon>0\). Choose \(\delta=\varepsilon\). For any \(z\in X\), if \(d(x,z)<\delta\), then \(d(\operatorname{id}_X(x),\operatorname{id}_X(z))=d(x,z)<\delta=\varepsilon\). So the identity map is epsilon-delta continuous at every point.
For an open set \(V\subseteq X\), the preimage is \(\operatorname{id}_X^{-1}(V)=\{x\in X:\operatorname{id}_X(x)\in V\}=V\). It is open because \(V\) is open. This verifies the open-set condition directly. The equality of the two sides in the distance calculation also checks that the same metric is being used in domain and target.
Worked Example: A Constant Map
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces, fix \(y_0\in Y\), and define \(f:X\to Y\) by \(f(x)=y_0\) for every \(x\in X\). Verify both conditions, including what happens to an arbitrary open set \(V\subseteq Y\).
Fix \(x\in X\) and \(\varepsilon>0\), and choose \(\delta=1\). Whenever \(d_X(x,z)<1\), the output distance is \(d_Y(f(x),f(z))=d_Y(y_0,y_0)=0<\varepsilon\). Thus \(f\) is continuous at every \(x\). The choice \(1\) is only a convenient positive radius; the output distance is zero regardless of the input distance.
If \(y_0\in V\), then every \(x\in X\) maps into \(V\), and \(f^{-1}(V)=X\). If \(y_0\notin V\), no \(x\) maps into \(V\), and \(f^{-1}(V)=\varnothing\). Both \(X\) and \(\varnothing\) are open in every topology, so every open set has open preimage. The two cases also cover \(V=\varnothing\) and \(V=Y\).
Worked Example: The Squaring Map
For \(f:\mathbb{R}\to\mathbb{R}\), \(f(x)=x^2\), prove pointwise epsilon-delta continuity and check a particular open-set preimage. Both copies of \(\mathbb{R}\) have their standard metric and topology.
Fix \(c\in\mathbb{R}\) and \(\varepsilon>0\). Choose \(\delta=\min\{1,\varepsilon/(1+2|c|)\}\), which is positive because \(1+2|c|>0\). If \(|x-c|<\delta\), the triangle inequality gives \(|x+c|=|(x-c)+2c|\le |x-c|+2|c|<1+2|c|\); the strict inequality follows from \(|x-c|<\delta\le1\). Therefore \[ |x^2-c^2|=|x-c||x+c| \le |x-c|(|x-c|+2|c|) <\delta(1+2|c|) \le\varepsilon. \] For the strict middle inequality, the second factor is less than \(1+2|c|\), and \(|x-c|<\delta\); if \(x=c\), the left side is zero and the displayed weak inequality still holds, while the final strict estimate follows directly from \(0<\delta(1+2|c|)\le\varepsilon\). Thus the required output distance is less than \(\varepsilon\).
For a numerical instance, take \(c=1\) and \(\varepsilon=1\). Then \(1+2|c|=1+2=3\), so \(\delta=\min\{1,1/3\}=1/3\). If \(|x-1|<1/3\), then \(|x+1|\le |x-1|+2<1/3+2=7/3\), and \(|x^2-1|=|x-1||x+1|<(1/3)(7/3)=7/9<1\). At \(x=1\), the actual output distance is \(0<1\), as required.
For an open-set check, let \(V=(1,4)\), an open subset of the target real line. The condition \(1<x^2<4\) holds exactly when \(-2<x<-1\) or \(1<x<2\). Hence \(f^{-1}(V)=(-2,-1)\cup(1,2)\), which is open: each point in either interval has a sufficiently small open interval around it contained in that same interval. This concrete preimage agrees with the theorem.
Worked Example: A Discontinuous Step Map
Define \(g:\mathbb{R}\to\mathbb{R}\) by \(g(x)=0\) when \(x\le0\), and \(g(x)=1\) when \(x>0\). Show directly that the map fails both continuity conditions at \(0\).
For the epsilon-delta test, take \(\varepsilon=1/2\). Given any \(\delta>0\), set \(x=\delta/2\). Then \(0<x\), so \(|x-0|=\delta/2<\delta\), but \(|g(x)-g(0)|=|1-0|=1\), which is not less than \(1/2\). Thus no \(\delta\) works at \(0\).
For the open-set test, take the open target interval \(V=(-1/2,1/2)\). The value \(0\) belongs to \(V\) and the value \(1\) does not, so \(g^{-1}(V)=(-\infty,0]\). This set is not open in \(\mathbb{R}\): every open ball around \(0\) contains positive real numbers, which are not in the set. Consequently the preimage condition fails as well. Both failures occur at the same boundary point.
A Common Pitfall and Why the Equivalence Helps
A common error in epsilon-delta estimates is to turn a weak inequality into a strict one without checking equality cases. In the squaring example, one might write \(|x+c|=|(x-c)+2c|\le |x-c|+2|c|<1+2|c|\) and then multiply by \(|x-c|\). The first inequality is valid, but the strict bound alone does not justify claiming that \(|x-c||x+c|<|x-c|(1+2|c|)\) when \(x=c\): that would assert \(0<0\), which is false.
The corrected argument keeps the weak inequality at the multiplication step: \[ |x^2-c^2|=|x-c||x+c| \le |x-c|(|x-c|+2|c|) <\delta(1+2|c|) \le\varepsilon. \] Here \(|x-c|<\delta\le1\) ensures the middle strict bound; if \(x=c\), the middle bound still compares zero with the positive quantity \(\delta(1+2|c|)\). This is why checking equality cases is part of a valid estimate, not a cosmetic detail.
The open-set formulation is useful because it packages all local distance tests into one rule about preimages. Conversely, the epsilon-delta formulation reveals what open preimages mean near a particular point. In applying the equivalence, keep the direction straight: continuity controls preimages of open target sets. It does not assert that images of open domain sets are open; the step map already shows that these are different kinds of statements.
Check Your Understanding
Use the definitions and the two directions of the theorem to answer the following.
- In the proof from pointwise continuity to open-set continuity, where is openness of \(V\) used, and where is continuity of \(f\) used?
- Suppose the preimage of every open ball in \(Y\) is open in \(X\). Explain why this suffices to show that the preimage of every open \(V\subseteq Y\) is open.
- For the constant map in the example, determine \(f^{-1}(V)\) separately when \(y_0\in V\) and when \(y_0\notin V\).
- For the squaring map, find the preimage of \((-1,1)\) and state whether it is open in \(\mathbb{R}\).
- For the step map \(g\), use the target open set \((1/2,3/2)\) to compute a preimage and decide whether that preimage is open.