The Timpson Lyceum · General Topology (Point Set Topology)

Continuity and the Epsilon-Delta Definition – Lecture 14 (Topology)

Overview

This lecture opens the chapter on continuity, the property that lets us carry topological structure from one space to another. Intuitively, a continuous function preserves proximity: points that are close in the domain stay close in the codomain. We begin with the calculus picture — the famous $\varepsilon$–$\delta$ definition — built up through a hands-on "interval game" that makes the definition feel inevitable rather than intimidating. We then meet the topological definition of continuity (the preimage of every open set is open) and verify it on concrete finite spaces and on the standard topology of $\mathbb{R}$. Along the way we prove that the identity, inclusion, and constant maps are continuous, reduce continuity-checking to basis elements, and finish with a closure characterization and a topological restatement of the $\varepsilon$–$\delta$ condition.

Key Concepts

Continuity, intuitively. A continuous function preserves proximity: if two points are close in the domain, their images are close in the codomain. The grade-school picture is "a function you can draw without lifting your pencil."

The $\varepsilon$–$\delta$ definition (calculus). A function $f:\mathbb{R}\to\mathbb{R}$ is continuous at $x_0$ if and only if $\forall\,\varepsilon>0\ \ \exists\,\delta>0\ \ \text{such that}\ \ |x-x_0|<\delta \implies |f(x)-f(x_0)|<\varepsilon.$ Read $|x-x_0|$ as the distance from $x$ to $x_0$. The condition is a game: the challenger picks any $\varepsilon$ (an output tolerance), and you must produce a $\delta$ (an input tolerance) so that every input within $\delta$ of $x_0$ lands within $\varepsilon$ of $f(x_0)$. If you can always answer, $f$ is continuous at $x_0$; if even one $\varepsilon$ defeats you, it is not.

Continuous at a point vs. continuous function. A function is continuous (as a whole) if it is continuous at every point of its domain. A single bad point breaks continuity of the whole function, even though the function may be continuous at infinitely many other points.

Why a function can fail the game. At a jump, points on either side of $x_0$ are close in the domain but their images are far apart. For a small enough $\varepsilon$, no $\delta$ works — proximity is destroyed, mirroring the "lift the pencil" picture.

The topological definition. Let $X,Y$ be topological spaces and $f:X\to Y$. Then $f$ is continuous if and only if for every open set $V\subseteq Y$, its preimage $f^{-1}(V)$ is open in $X$. Mnemonic: the preimage of open sets is open. Note $f$ need not be one-to-one or onto to be continuous.

Useful preimage identities. For sets $A,B$:

• $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$
• $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$
• $f^{-1}(A\setminus B)=f^{-1}(A)\setminus f^{-1}(B)$

These extend to arbitrary unions, which is what makes the basis reduction below work.

Standard maps that are continuous.

• Identity $\mathrm{id}:X\to X$, $\mathrm{id}(x)=x$: for any open $U$, $f^{-1}(U)=U$, which is open.
• Constant $f:X\to Y$, $f(x)=b$ for a fixed $b\in Y$: for any open $V$, $f^{-1}(V)=X$ if $b\in V$ and $f^{-1}(V)=\varnothing$ if $b\notin V$; both are open.
• Inclusion $i:Y\hookrightarrow X$ of a subspace $Y\subseteq X$, $i(y)=y$: for open $U\subseteq X$, $i^{-1}(U)=U\cap Y$, which is exactly an open set in the subspace topology.

Basis reduction (key labor-saving theorem). Let $\beta$ be a basis for the topology on $Y$. Then $f:X\to Y$ is continuous if and only if $f^{-1}(B)$ is open in $X$ for every basis element $B\in\beta$. You need only check basis elements, not all open sets — because every open $V$ is a union of basis elements $V=\bigcup_\alpha B_\alpha$, and $f^{-1}(V)=\bigcup_\alpha f^{-1}(B_\alpha)$ is then a union of open sets, hence open.

Continuous maps and closures. If $f:X\to Y$ is continuous and $A\subseteq X$, then $x\in\overline{A}\implies f(x)\in\overline{f(A)}$. Equivalently, $f(\overline{A})\subseteq\overline{f(A)}$. (Caution: $\overline{f(A)}$ is generally larger than $f(A)$ — this is not the same as $f$ of the closure being equal to the closure of $f(A)$.) The proof uses the neighborhood criterion $x\in\overline{A}\iff$ every neighborhood of $x$ meets $A$, applied in its contrapositive form.

Topological restatement of $\varepsilon$–$\delta$. $f:X\to Y$ is continuous if and only if for every $x$ and every open $V\subseteq Y$ with $f(x)\in V$, there exists a neighborhood $U_x$ of $x$ with $f(U_x)\subseteq V$. This is the $\varepsilon$–$\delta$ idea with $V$ playing the role of the $\varepsilon$-interval and $U_x$ the $\delta$-interval — proven (next time) to be equivalent to the preimage-of-open-sets definition.

Worked Example

Goal. Use the basis reduction to prove $f(x)=2x+1$ is continuous on $\mathbb{R}$ (standard topology, basis = open intervals). It suffices to show the preimage of an arbitrary open interval $(a,b)$ is open.

Solve $f(x)=y$ for $x$ to find the preimage of a generic basis element:

The preimage of every open interval is again an open interval — an open set — so by the basis theorem $f$ is continuous.

Sanity check with the game at $x_0=2$ (so $f(2)=5$). Because $f$ doubles distances, choosing $\delta=\varepsilon/2$ always works:

For any challenge $\varepsilon$, the response $\delta=\varepsilon/2$ forces every output within $\varepsilon$ of $5$, so $f$ is continuous at $2$.

Summary

• A continuous function preserves proximity; the grade-school version is "drawable without lifting the pencil."
• The $\varepsilon$–$\delta$ definition is a game: for every output tolerance $\varepsilon$ you must find an input tolerance $\delta$ with $|x-x_0|<\delta\Rightarrow|f(x)-f(x_0)|<\varepsilon$; one losing $\varepsilon$ means discontinuity at that point.
• A function is continuous overall only if it is continuous at every point of its domain.
• The topological definition: $f$ is continuous iff $f^{-1}(V)$ is open for every open $V$ — the preimage of open sets is open (no injectivity/surjectivity required).
• Identity, constant, and inclusion maps are continuous; checking continuity reduces to verifying preimages of basis elements are open.
• Continuity respects closures, $f(\overline{A})\subseteq\overline{f(A)}$, and the $\varepsilon$–$\delta$ condition restates topologically as: for each $f(x)\in V$ open, some neighborhood $U_x$ has $f(U_x)\subseteq V$.