Equivalent Formulation of Continuity¶

Theorem¶

Let $(X,d_{X})$ and $(Y,d_{Y})$ be metric spaces, and let $f\colon X\to Y$ be a function. The following statements are equivalent:

• $f$ is continuous.
• For every $x_0\in X$ and every convergent sequence $(x_i)_{i=1}^\infty$ in $(X,d_{X})$ converging to $x_0$, its image under $f$, $f(x_i)_{i=1}^\infty$, also converges to $f(x_0)$.
• The preimage of any open set in $(Y,d_{Y})$ is open in $(X,d_{X})$.
• The preimage of any closed set in $(Y,d_{Y})$ is closed in $(X,d_{X})$.

Interpretation¶

The theorem gives three tests for continuity that avoid $\varepsilon$ and $\delta$. A function is continuous if and only if it preserves limits of sequences, and if and only if the preimage operation preserves openness, or equivalently closedness.

In $X$ the points $x_i$ approach $x_0$, and in $Y$ their images $f(x_i)$ approach $f(x_0)$: this is the sequential condition. The dashed sets show the open-set condition. $V$ is an open set around $f(x_0)$, and its preimage $f^{-1}(V)$, the set of all points that $f$ sends into $V$, is an open set around $x_0$. The two conditions fit together: since $x_i\to x_0$ and $f^{-1}(V)$ is open, the $x_i$ eventually lie in $f^{-1}(V)$, so the $f(x_i)$ eventually lie in $V$.

The conditions concern preimages, not images: a continuous function need not send open sets to open sets, as the Examples show. Preimages behave well because they respect complements, $f^{-1}(Y\setminus V)=X\setminus f^{-1}(V)$$f^{-1}(V)$, which is why the open-set and closed-set versions are equivalent.

One sequence $x_i\to x_0$ with $f(x_i)\not\to f(x_0)$ proves discontinuity. The preimage forms prove that sets are open or closed: the open unit disc $\{(x,y)\in\mathbb{R}^2\colon x^2+y^2<1\}$ is the preimage of the open set $(-\infty,1)$ under the continuous $f(x,y)=x^2+y^2$$y^2$, so it is open, and every level set of a continuous real-valued function is closed.

The open-set form uses no distances at all. That is why it serves as the definition of continuity in a topological space, and why replacing a metric by a topologically equivalent one, which has the same open sets, does not change which functions are continuous.

Key Results¶

• At a fixed point $x_0\text{,}$ the sequential condition is equivalent to continuity at $x_0\text{.}$ In metric spaces it is often the easiest to work with when proving discontinuity. To show $f$ is not continuous at a point, find one sequence converging to that point whose image does not converge to the function value.

Examples¶

• A continuous function $f$ might map an open set $V$ to a non-open set $f(V)$. For example, let $V=(0,1)$ with $f(x)=1$ for all $x\in\mathbb{R}$$\mathbb{R}$. Then $f(V)=\{1\}$ isn't open under usual metrics.
• A continuous function $f$ might map a closed set $F$ to a non-closed set $f(F)$. For example, let $F=[1,+\infty)$ and $f(x)=1/x$ for all $x>0$$0$. Then $f(F)=(0,1]$ isn't closed under usual metrics.

Prerequisites¶

• Continuity between Metric Spaces
• Convergence in a Metric Space
• Image and Preimage of a Set
• Open and Closed Sets in a Metric Space

Related Concepts¶

• Uniform Continuity between Metric Spaces
• Topology and Topological Space
• Topologically Equivalent Metrics

Sources¶

• Math III: Analysis · Lecture 8

Aliases: equivalent definition of continuity, sequential criterion for continuity, sequential characterization of continuity Symbols: delta, element of, epsilon, infinity, real numbers

Last updated: 20 October 2026 03:02