Beyond Sequences: A Topological Approach to Density Arguments

Every time we open a textbook on real analysis or functional analysis, proofs about density invariably begin with a familiar pattern: "Since $M$$M$ is dense in $X$$X$, for any $x\in X$$x∈X$, there exists a sequence $xₙ\subseteq M$${xₙ}⊆M$ such that $xₙ\to x...$$xₙ→x...$ This sequence construction has become almost reflexive, to the point where we forget to ask: Is this really the best way to handle density?

This sequence-based approach has multiple issues. First, in infinite spaces, constructing such sequences often relies on the axiom of choice, and this non-constructive nature robs the proof of its desired transparency. Second, while this method remains valid in general topological spaces, it overly depends on our intuition of "approximation" developed in metric spaces. When we encounter generic points in algebraic geometry, this intuition completely breaks down: a set consisting of a single point can be dense! Here, any sequence "converging" to this point must be trivially constant, clearly indicating that we shouldn't understand density through sequence limits.

More importantly, these sequence arguments obscure the essential topological meaning of density: a subset is dense if and only if its closure is the whole space. This simple definition suggests that when dealing with density, we should be thinking about set inclusion and closure properties rather than laboring to construct sequences. This way of thinking is not only more essential but maintains intuitive consistency across all topological spaces.

This article will introduce a unified method based on pure topology, which not only elegantly resolves traditional problems but also helps us develop a more fundamental topological intuition...

Proposition. Let $f:X\to Y$$f:X\to Y$ be a continuous function and $M$$M$ be a dense subset of $X$$X$, $N$$N$ is a closed subset of $Y$$Y$. If $f(M)\subseteq N$$f(M)\subseteq N$ then $f(X)\subseteq N$$f(X)\subseteq N$.

Proof. Observe that $N$$N$ is closed implies that $f^{-1}(N)$$f^{-1}(N)$ is closed as well. Also we have $M\subseteq f^{-1}(N)$$M\subseteq f^{-1}(N)$ hence $f^{-1}(N)$$f^{-1}(N)$ is a closed subset contain $M$$M$. $M$$M$ is dense implies that $f^{-1}(N)=X$$f^{-1}(N)=X$. $◻$$\square$​

Corollary. Let $f:X\to Y$$f:X\to Y$ be a continuous function and $M$$M$ be a subset of $X$$X$, $N$$N$ is a closed subset of $Y$$Y$.

If $f(M)\subseteq N$$f(M)\subseteq N$ then $f(\mathrm{Cl}(M))\subseteq N$$f(\mathrm{Cl}(M))\subseteq N$.

Application.

1. In affine space

Zariski Dense Argument in Linear Algebra

2. In Hillbert space

Let $K$$K$ be a dense subset of $\mathcal{H}$$\mathcal H$ where $\mathcal{H}$$\mathcal H$ is a Hillbert space, then $K^{\mathrm{\perp }}=0$$K^{\bot}=0$.

Proof. Assume that $x\in K^{\mathrm{\perp }}$$x\in K^{\bot}$, then $f_x=⟨-,x⟩$$f_x=\langle-,x\rangle$ is continuous. Notice that Hillbert space is Hausdorff space hence $0$$0$ is a closed set. $K\subseteq f_x^{-1}(0)$$K\subseteq f_x^{-1}(0)$ and $K$$K$ is dense implies that $f_x^{-1}(0)=\mathcal{H}$$f^{-1}_x(0)=\mathcal H$, hence $x=0.$$x=0.$ $◻$$\square$

3. In Hausdorff Space

Let $f:X\to Y$$f:X\to Y$ be a continous function, $Y$$Y$ is Hausdorff and $M\subseteq X$$M\subseteq X$ is dense subset. If $f{|}_M=g{|}_M$$f|_M=g|_M$, then $f=g$$f=g$.

Proof. Notice that $Y$$Y$ is Hausdorff iff $\mathrm{\Delta }=\{(y,y)\in Y\times Y:y\in Y\}$$\Delta=\{(y,y)\in Y\times Y:y\in Y\}$​ is closed.

Hence the subset of $X$$X$, $E(f,g)=\{x\in X:f(x)=g(x)\}=(f,g{)}^{-1}\mathrm{\Delta }$$E(f,g)=\{x\in X:f(x)=g(x)\}=(f,g)^{-1}\Delta$ is closed as well.

Since $f{|}_M=g{|}_M$$f|_M=g|_M$, $M\subseteq E(f,g)$$M\subseteq E(f,g)$, $M$$M$ is dense implies that $E(f,g)=X$$E(f,g)=X$. $◻$$\square$

4. In space with generaic point.

Let $X$$X$ be a space with generaic point, i.e. there exists a point $x\in X$$x\in X$, $\mathrm{Cl}(\{x\})=X$$\mathrm{Cl}(\{x\})=X$. For example, $\mathrm{Spec}(\mathbb{Z})$$\mathrm{Spec}(\mathbb Z)$.

Let $f:X\to Y$$f:X\to Y$ be a continuous function and $x\in X$$x\in X$ be the generic point, if $f(x)$$f(x)$ is closed, then $f$$f$ is constant map.

Proof. $f(x)$$f(x)$ is closed implies that $f^{-1}(x)$$f^{-1}(x)$ is closed as well, but $\{x\}\subseteq f^{-1}(x)$$\{x\} \subseteq f^{-1}(x)$ hence $f^{-1}(x)=X$$f^{-1}(x)=X$. In particular, if $Y$$Y$ is Hausdorff then $f$$f$ has to be constant map. $◻$$\square$

Proposition. Let $Y\subseteq X$$Y\subseteq X$ be a connected set, then the closure of $Y$$Y$ is connected.

Proof.

A topological space $Z$$Z$ is disconnected iff there exits a continuous surjection $f:Z\twoheadrightarrow \{0,1\}$$f:Z\twoheadrightarrow \{0,1\}$ with discrete topology.

Hence let us consider a continuous function $f:\mathrm{Cl}(Y)\to \{0,1\}$$f:\mathrm{Cl}(Y)\to\{0,1\}$, then $f{|}_Y$$f|_Y$ is constant function since $Y$$Y$ is connected.

Let $f{|}_Y=1$$f|_Y=1$​, then $Y\subseteq f^{-1}(1)$$Y\subseteq f^{-1}(1)$ and $Y$$Y$ is dense in $\mathrm{Cl}(Y)$$\mathrm{Cl}(Y)$. Hence $f$$f$ has to be constant function. $◻$$\square$