From a Goal to a Metric Estimate
“Equivalent Metrics” showed how to organize proofs around balls, sequences, and estimates. This tutorial applies that style more broadly. In a metric-space proof, the main challenge is often not finding a complicated theorem; it is choosing the right point, sequence, or inequality so that the definition gives exactly what the conclusion requires.
A useful habit is to begin at the desired conclusion and work backward. If the goal is to prove \(u_n\to u\), identify the distance \(d(u_n,u)\) and ask how to make it smaller than an arbitrary \(\varepsilon>0\). If the goal is to prove that a point lies in a closure, try to construct points of the set within distance \(1/n\). If the goal is to disprove a universal claim, identify which choices are allowed to depend on which quantities.
For example, to prove \(d(a,c)<\varepsilon\) by splitting a path through \(b\), the triangle inequality gives \(d(a,c)\leq d(a,b)+d(b,c)\). It is enough to arrange \(d(a,b)<\varepsilon/2\) and \(d(b,c)<\varepsilon/2\). The strict inequalities matter: together they give \(d(a,c)<\varepsilon\), not merely \(d(a,c)\leq\varepsilon\).
Dense Data Determine Continuous Maps
A dense subset comes arbitrarily close to every point. Continuity allows that closeness in the domain to be transferred to closeness of function values. One consequence is that continuous maps cannot differ at a point if they agree on a dense subset.
Proof. Fix \(x\in X\). Since \(D\) is dense, every ball centered at \(x\) meets \(D\). Thus, for each positive integer \(n\), choose \(d_n\in D\) such that \(d_X(d_n,x)<1/n\). It follows that \(d_n\to x\). By continuity, \(f(d_n)\to f(x)\) and \(g(d_n)\to g(x)\). But \(f(d_n)=g(d_n)\) for every \(n\). The common sequence therefore converges both to \(f(x)\) and to \(g(x)\). Uniqueness of metric limits gives \(f(x)=g(x)\). Since \(x\) was arbitrary, the functions agree on all of \(X\). \(\square\)
The proof uses a specific sequence built from density, then continuity, then uniqueness of limits. It is important that \(f\) and \(g\) are both continuous: continuity is what identifies the limits of their values with their values at \(x\).
Worked Example: Recovering a Continuous Function from Rational Values
Suppose \(f:\mathbb{R}\to\mathbb{R}\) is continuous and \(f(q)=q^2\) for every rational number \(q\). The rationals are dense in \(\mathbb{R}\), and the function \(g(x)=x^2\) is continuous. The two functions agree on the dense subset \(\mathbb{Q}\). By Dense-Set Determination, $$ f(x)=g(x)=x^2 \quad\text{for every }x\in\mathbb{R}. $$
This conclusion does not require choosing a special rational near each real number separately. The theorem packages that choice into the sequence \(q_n\in\mathbb{Q}\) with \(|q_n-x|<1/n\). Continuity then gives \(f(q_n)\to f(x)\), while \(q_n^2\to x^2\), so the prescribed rational values force the value at \(x\).
Continuity Transfers Density Through a Map
The same sequence construction gives another useful result. If \(D\) is dense in \(X\) and \(f\) is continuous, then values of \(f\) on \(D\) approximate every value of \(f\) on \(X\). The image \(f(D)\) need not equal \(f(X)\), but it is dense in that image.
Proof. Since \(D\subseteq X\), we have \(f(D)\subseteq f(X)\). Monotonicity of closure gives \(\overline{f(D)}\subseteq\overline{f(X)}\).
For the reverse inclusion, fix \(x\in X\). Density gives a sequence \((d_n)\) in \(D\) such that \(d_n\to x\). Continuity of \(f\) gives \(f(d_n)\to f(x)\). Each \(f(d_n)\) belongs to \(f(D)\), so the Sequential Characterization of Closure shows that \(f(x)\in\overline{f(D)}\). Hence \(f(X)\subseteq\overline{f(D)}\). Because \(\overline{f(D)}\) is closed and contains \(f(X)\), it contains \(\overline{f(X)}\). Thus \(\overline{f(X)}\subseteq\overline{f(D)}\), proving equality. \(\square\)
The conclusion concerns closure in the target space, not equality of the two images. Also, it says that \(f(D)\) is dense in \(f(X)\) with its relative metric: its closure in \(Y\) contains \(f(X)\). It does not claim that \(f(X)\) is closed.
Worked Example: Squaring Rational Points in an Interval
Let \(X=(-1,1)\), let \(D=\mathbb{Q}\cap(-1,1)\), and define \(f(x)=x^2\). The set \(D\) is dense in \(X\), and \(f\) is continuous. The image of the full interval is $$ f(X)=[0,1). $$ The theorem therefore gives \(\overline{f(D)}=\overline{[0,1)}=[0,1]\), with closure taken in \(\mathbb{R}\). In particular, rational squares from this interval approximate every number in \([0,1)\), and their closure also contains the endpoint \(1\), which is not in \(f(X)\).
The endpoint illustrates why the theorem is stated using closures. The image \(f(X)\) itself is not closed in \(\mathbb{R}\), whereas the values \(f(q)\) with \(q\in D\) are dense in its closure.
Checking the Hypotheses: A Necessary Caution
A proof should use every hypothesis at the point where it is needed. In the density theorem, density supplies a sequence approaching \(x\), while continuity ensures that the image sequence approaches \(f(x)\). If continuity is removed, the image of a dense set may fail even to approximate some values in the full image.
Worked Example: A Dense Set Whose Image Is Not Dense
Take \(X=Y=\mathbb{R}\), let \(D=\mathbb{Q}\), and define $$ h(x)= \begin{cases} 1,&x\in\mathbb{Q},\\ 0,&x\notin\mathbb{Q}. \end{cases} $$ Then \(D\) is dense in \(X\), but \(h(D)=\{1\}\), whose closure is \(\{1\}\). On the other hand, \(h(X)=\{0,1\}\), so \(\overline{h(X)}=\{0,1\}\). Therefore \(\overline{h(D)}\ne\overline{h(X)}\). The missing hypothesis is continuity: for example, every ball around any real number contains both rationals and irrationals, so \(h\) is not continuous at any point.
This example is also a model for testing whether a theorem’s assumptions are essential. Remove one hypothesis, compute both sides of the proposed conclusion, and check whether they still agree. A valid counterexample does more than show that a proof cannot proceed; it shows that the claim itself can fail.
A Reliable Workflow for Metric Proofs
Several recurring choices make proofs more efficient. When a conclusion concerns every point, fix an arbitrary point first. When density is available, turn it into a sequence with distances less than \(1/n\). When continuity is available, apply it to that sequence. When a proof needs convergence, write the relevant target distance rather than relying on an informal description of points being “close.”
Identify the point, distance, or set-membership statement that must be established.
Use a radius for a ball argument, a sequence for a sequential argument, or a specific pair of points for a counterexample.
For density arguments, construct the approaching sequence first; apply continuity only after that sequence has been identified.
State the estimate or theorem that turns the construction into the desired conclusion, such as uniqueness of limits or the sequential characterization of closure.
A common pitfall is to reverse quantifiers. Continuity at a point lets the input radius depend on the desired output tolerance and the point. Uniform continuity requires a single input radius for all points. Likewise, density lets the approximating point depend on the center and the requested accuracy. Keeping track of these dependencies prevents an argument from claiming more than its hypotheses provide.
Another useful check is to distinguish a set from its closure. A convergent sequence of values in \(f(D)\) shows that its limit belongs to \(\overline{f(D)}\); it does not show that the limit belongs to \(f(D)\). In metric spaces, the Sequential Characterization of Closure makes this distinction precise. Similar care with “less than” versus “less than or equal to” keeps epsilon estimates valid at the final step.
Check Your Understanding
Use the proof strategies and results in this tutorial to answer the following questions.
- In Dense-Set Determination, where is density used, and where is continuity used?
- Why does a sequence of points in \(D\) approaching \(x\) exist when \(D\) is dense?
- What does the continuous-image density theorem say about \(\overline{f(D)}\) and \(\overline{f(X)}\)?
- Why does that theorem not assert that \(f(D)=f(X)\)?
- What fails in the example with \(h(x)=1\) on rationals and \(h(x)=0\) on irrationals?
- In an epsilon estimate using the triangle inequality, why might one split \(\varepsilon\) into two smaller positive tolerances?