Topology: Limits, Compactness, Continuity, and Correspondences
Y. Eddie Lu, Summer 2026
ECON 8001 course index · Lecture 4 of 8
Orientation
What problem does topology solve?
Topology supplies the language for claims such as “nearby inputs have nearby outputs” and “an optimizing sequence has a feasible limit.” It is where calculus gains its hypotheses: a maximum may fail to exist because a feasible set has a missing boundary point, and a correspondence may fail to be continuous because a whole set of choices changes abruptly.
Where this sits
Topology uses the spaces and maps from Linear algebra: objects and supplies the compactness and continuity hypotheses used in Differentiation and Optimization.
All ordinary vectors are columns in \(\mathbb R^n\) and use the Euclidean norm. For a set \(X\subseteq\mathbb R^n\), topology is relative to the stated ambient space unless said otherwise. A sequence has changing terms \(x_k\) and a fixed candidate limit \(x^*\); do not use the same symbol for both roles.
Prerequisite retrieval. Is \([0,1)\) open in \(\mathbb R\)?
No. The included point \(0\) has no open interval contained in \([0,1)\), and the omitted limit point \(1\) means the set is not closed either. It is open only relative to the subspace \([0,1]\).
1. Distance, balls, and set geometry
For \(x=(x_1,\ldots,x_n)^{\mathsf{T}}\in\mathbb R^n\), define \[\|x\|=\left(\sum_{i=1}^n x_i^2\right)^{1/2}.\] For \(x,y\in\mathbb R^n\), the Euclidean metric is \[d(x,y)=\|x-y\|. \] It measures distance and obeys nonnegativity, symmetry, and the triangle inequality, with \(d(x,y)=0\) if and only if \(x=y\).
For \(x_0\in\mathbb R^n\) and \(r>0\), the open ball is \[B_r(x_0)=\{x\in\mathbb R^n:\|x-x_0\|<r\}.\] If \(X\subseteq\mathbb R^n\), a point \(x_0\in X\) is an interior point of \(X\) when \(B_r(x_0)\subseteq X\) for some \(r>0\). The interior \(\operatorname{int}X\) is the set of interior points. The set \(X\) is open if every one of its points is interior. A set \(Y\) is a neighborhood of \(X\) when \(X\subseteq\operatorname{int}Y\).
In \(\mathbb R\), \([0,1)\) is neither open nor closed. In the subspace \([0,1]\), it is open: for each \(x\in[0,1)\), choose \(r>0\) with \(x+r<1\); then \(B_r(x)\cap[0,1]\subseteq[0,1)\). At \(x=0\), for example, the relative ball is \([0,r)\). Always name the ambient space before classifying a set.
A point \(x\in\mathbb R^n\) is a limit point of \(X\subseteq\mathbb R^n\) if every ball \(B_r(x)\) with \(r>0\) contains a point of \(X\) distinct from \(x\). The closure \(\overline X\) is \(X\) together with all of its limit points. The boundary is \[\partial X=\overline X\setminus\operatorname{int}X.\] A boundary point is one whose every ball meets both \(X\) and its complement.
A set \(X\subseteq\mathbb R^n\) is closed if it contains all of its limit points. Equivalently in \(\mathbb R^n\), \(X\) is closed if its complement is open, or if \(X=\overline X\).
The set \((0,1)\) is bounded but not closed in \(\mathbb R\): the sequence \(1/(k+1)\) lies in the set and converges to \(0\notin(0,1)\). The missing point is why boundedness alone cannot produce a feasible limit.
2. Sequences make limits testable
A sequence in a set \(X\) is a function from \(\mathbb N\) to \(X\), written \((x_k)_{k\in\mathbb N}\). A subsequence is \((x_{k_j})_{j\in\mathbb N}\) where \(k_1<k_2<\cdots\) are positive integers. A sequence in \(\mathbb R^n\) is bounded if there is \(M<\infty\) such that \(\|x_k\|\le M\) for every \(k\).
The sequence converges to \(x^*\in\mathbb R^n\), written \(x_k\to x^*\), if:
- for every \(\varepsilon>0\),
- there exists \(K\in\mathbb N\) s.t.,
- for every \(k\ge K\), we have \(\|x_k-x^*\|<\varepsilon\).
For \(x_k=1/k\), given \(\varepsilon>0\), choose any integer \(K>1/\varepsilon\). Then \(k\ge K\) implies \(|x_k-0|=1/k\le1/K<\varepsilon\). Thus \(x_k\to0\). The index \(k\) changes; the limit \(0\) does not.
- A convergent sequence in \(\mathbb R^n\) has a unique limit.
- Every subsequence of a sequence converging to \(x^*\) also converges to \(x^*\).
- For \(x_k=(x_{k,1},\ldots,x_{k,n})^{\mathsf{T}}\) and \(x^*=(x_1^*,\ldots,x_n^*)^{\mathsf{T}}\), \(x_k\to x^*\) if and only if \(x_{k,i}\to x_i^*\) for every \(i=1,\ldots,n\).
Proof strategy. For uniqueness, suppose a sequence converges to \(x^*\) and \(y^*\) and use the triangle inequality with two balls of radius \(\varepsilon/2\). For subsequences, an increasing index eventually exceeds the original \(K\). For componentwise convergence, bound the Euclidean norm by the coordinate errors, and conversely use \(|x_{k,i}-x_i^*|\le\|x_k-x^*\|\).
Every bounded sequence in \(\mathbb R^n\) has a convergent subsequence.
Proof strategy. In one dimension, repeatedly bisect a bounded interval and retain a half containing infinitely many terms; the nested intervals identify a convergent subsequence. In \(\mathbb R^n\), apply the one-dimensional result coordinate by coordinate and retain subsequences. Boundedness is the
assumption that starts this extraction.
For \(X\subseteq\mathbb R^n\), the following are equivalent:
- \(X\) is closed.
- Every convergent sequence \((x_k)\) with \(x_k\in X\) for all \(k\) has its limit in \(X\).
Proof strategy. If \(X\) contains all its limit points, any limit of a sequence from \(X\) is either one of its terms or a limit point. Conversely, if \(X\) omits a limit point \(x\), choose a point of \(X\setminus\{x\}\) from each ball \(B_{1/k}(x)\); the resulting sequence converges to \(x\) and violates the sequential condition.
Retrieval check. Why does the alternating sequence \((-1)^k\) fail to converge?
Its even subsequence converges to \(1\) and its odd subsequence to \(-1\). If the original sequence converged, both subsequences would have the same limit by Theorem 2.3.6.
3. Compactness retains limit points
In \(\mathbb R^n\), a set \(X\) is compact if it is closed and bounded. More generally, compactness means that every open cover of \(X\) has a finite subcover. The closed-and-bounded test is special to finite-dimensional Euclidean space.
For \(X\subseteq\mathbb R^n\), the following are equivalent:
- \(X\) is compact.
- Every sequence in \(X\) has a subsequence converging to an element of \(X\).
Proof strategy. If \(X\) is closed and bounded, Bolzano–Weierstrass gives a convergent subsequence and closedness keeps its limit in \(X\). Conversely, failure of boundedness creates a sequence whose norms diverge; failure of closedness creates a sequence approaching a missing limit point. Either failure contradicts the sequential property.
\([0,1]\) is compact. \((0,1)\) is bounded but not compact because it omits \(0\) and \(1\). The set \(\{0\}\cup\{1/k:k\in\mathbb N\}\) is compact: it is bounded and it includes its only nontrivial limit point, \(0\).
To show that an extremum is attained, start with a maximizing or minimizing sequence. Compactness produces a convergent subsequence whose limit remains feasible. A continuity or semicontinuity condition then transfers the limiting objective value to that feasible limit. Compactness alone does not compare objective values.
4. Continuity and one-sided continuity of functions
Let \(X\subseteq\mathbb R^n\) and \(f:X\to\mathbb R^m\). The function is continuous at \(x_0\in X\) if:
- for every \(\varepsilon>0\),
- there exists \(\delta>0\) s.t.,
- for every \(x\in X\):
- if \(\|x-x_0\|<\delta\),
- then \(\|f(x)-f(x_0)\|<\varepsilon\).
It is continuous on \(X\) if it is continuous at every \(x_0\in X\).
For \(f:X\to\mathbb R^m\) and \(x_0\in X\), \(f\) is continuous at \(x_0\) if and only if every sequence \(x_k\in X\) satisfying \(x_k\to x_0\) also satisfies \(f(x_k)\to f(x_0)\).
Proof strategy. The forward implication substitutes a convergent sequence into the \(\varepsilon\)–\(\delta\) definition. For the reverse implication, negate continuity: if a fixed \(\varepsilon\) defeats every \(\delta\), select \(x_k\) within \(1/k\) of \(x_0\) whose image remains at least \(\varepsilon\) away.
Let \(a<b\), let \(f:[a,b]\to\mathbb R\) be continuous, and let \[y\in[\min\{f(a),f(b)\},\max\{f(a),f(b)\}].\] Then there exists \(c\in[a,b]\) such that \[f(c)=y.\]
The conclusion is \(f(c)=y\), not \(f(c)=c\). The point \(c\) is an input chosen to hit the target output level \(y\).
Proof strategy. Suppose \(f(a)\le y\le f(b)\) and let \(S=\{x\in[a,b]:f(x)\le y\}\). Completeness gives \(c=\sup S\). Continuity rules out both \(f(c)<y\) and \(f(c)>y\), so \(f(c)=y\). Reverse the inequalities when \(f(b)\le y\le f(a)\).
Semicontinuity
Let \(f:X\to[-\infty,\infty]\) and \(x_0\in X\). The function is upper semicontinuous (usc) at \(x_0\) if every sequence \((x_k)\) in \(X\) with \(x_k\to x_0\) satisfies \[\limsup_{k\to\infty}f(x_k)\le f(x_0).\] It is lower semicontinuous (lsc) at \(x_0\) if every sequence \((x_k)\) in \(X\) with \(x_k\to x_0\) satisfies \[\liminf_{k\to\infty}f(x_k)\ge f(x_0).\]
Upper semicontinuity forbids an upward jump at \(x_0\) but permits a downward jump. Lower semicontinuity forbids a downward jump but permits an upward jump. That is why usc fits maximization and lsc fits minimization.
A real-valued function is continuous if and only if it is both upper semicontinuous and lower semicontinuous.
Proof strategy. If \(f(x_k)\to f(x_0)\), then its limsup and liminf both equal \(f(x_0)\). Conversely, the two inequalities force limsup and liminf of \(f(x_k)\) to agree with \(f(x_0)\), which gives sequential continuity.
Let \(f(0)=1\) and \(f(x)=0\) for \(x\ne0\). At \(0\), \(f\) is usc but not lsc: nearby values do not rise above \(1\), but they fall to \(0\). This is a statement about a numerical function, not about a correspondence.
Uniform continuity
For \(f:X\to\mathbb R^m\), uniform continuity means:
- for every \(\varepsilon>0\),
- there exists \(\delta>0\) s.t.,
- for every \(x,y\in X\):
- if \(\|x-y\|<\delta\),
- then \(\|f(x)-f(y)\|<\varepsilon\).
The chosen \(\delta\) may depend on \(\varepsilon\) but not on the location in \(X\). Ordinary continuity allows \(\delta\) to depend on \(x_0\).
The function \(f(x)=1/x\) is continuous on \((0,1)\) but not uniformly continuous there. Let \(x_k=1/(k+1)\) and \(y_k=1/(k+2)\). Then \(|x_k-y_k|\to0\), while \[|f(x_k)-f(y_k)|=|(k+1)-(k+2)|=1.\] The domain lets inputs approach the missing boundary point \(0\), where the slope becomes unbounded.
If \(X\subseteq\mathbb R^n\) is compact and \(f:X\to\mathbb R^m\) is continuous, then \(f\) is uniformly continuous.
Proof strategy. If uniform continuity failed, choose pairs \(x_k,y_k\in X\) with \(\|x_k-y_k\|<1/k\) but outputs separated by a fixed \(\varepsilon>0\). Compactness gives a subsequence \(x_{k_j}\to x^*\in X\). The paired points also converge to \(x^*\) by the triangle inequality, so continuity makes the output distance tend to zero, a contradiction.
Retrieval check. Which is stronger on a noncompact domain: continuity or uniform continuity?
Uniform continuity. It requires one \(\delta\) for every pair of points in the domain; ordinary continuity chooses a possibly different \(\delta\) at each center point.
5. Correspondences and hemicontinuity
Let \(X\subseteq\mathbb R^n\) and \(Y\subseteq\mathbb R^m\). A correspondence \(\varphi:X\rightrightarrows Y\) assigns to each \(x\in X\) a subset \(\varphi(x)\subseteq Y\). It is nonempty-valued, open-valued, closed-valued, or compact-valued when every value \(\varphi(x)\) is respectively nonempty, open in \(Y\), closed in \(Y\), or compact.
A function is the special case in which each \(\varphi(x)\) is a singleton.
The correspondence \(\varphi\) is upper hemicontinuous (uhc) at \(x_0\in X\) if, for every open \(V\subseteq Y\) with \(\varphi(x_0)\subseteq V\), there is a neighborhood \(U\) of \(x_0\) such that \[x\in U\cap X\Longrightarrow\varphi(x)\subseteq V.\]
The correspondence \(\varphi\) is lower hemicontinuous (lhc) at \(x_0\in X\) if, for every open \(V\subseteq Y\) with \(V\cap\varphi(x_0)\ne\varnothing\), there is a neighborhood \(U\) of \(x_0\) such that \[x\in U\cap X\Longrightarrow V\cap\varphi(x)\ne\varnothing.\] It is continuous at \(x_0\) if it is both uhc and lhc there.
Uhc draws an open fence around the whole value set \(\varphi(x_0)\) and requires all nearby value sets to remain inside. Lhc selects an open region that touches \(\varphi(x_0)\) and requires nearby value sets to keep touching that region. These are statements about changing sets, not the one-sided inequalities that define semicontinuity of a function.
Let \(\varphi(0)=\{0\}\) and \(\varphi(x)=\{0,1\}\) for \(x\ne0\). Then \(\varphi\) is not uhc at \(0\): the remote point \(1\) appears arbitrarily nearby; lhc holds.
Let \(\psi(0)=\{0,1\}\) and \(\psi(x)=\{0\}\) for \(x\ne0\). Then \(\psi\) is not lhc at \(0\): the choice \(1\) disappears; uhc holds.
Let \(\varphi:X\rightrightarrows Y\) be upper hemicontinuous and compact-valued. If \(x_k\to x^*\) in \(X\), \(y_k\to y^*\) in \(Y\), and \(y_k\in\varphi(x_k)\) for every \(k\), then \(y^*\in\varphi(x^*)\).
Proof. Suppose instead that \(y^*\notin\varphi(x^*)\). Since \(\varphi(x^*)\) is compact, its distance from \(y^*\) is positive. Choose \(0<r<\operatorname{dist}(y^*,\varphi(x^*))\) and let
\[V=\{y\in Y:\operatorname{dist}(y,\varphi(x^*))<r\}.\]
This set is open in \(Y\), contains \(\varphi(x^*)\), and has closure disjoint from a neighborhood of \(y^*\). Upper hemicontinuity gives \(\varphi(x_k)\subseteq V\) eventually, hence \(y_k\in V\) eventually. But \(y_k\to y^*\) eventually places \(y_k\) in that disjoint neighborhood, a contradiction.
Compact images under upper hemicontinuity
Let \(\varphi:X\rightrightarrows Y\) be compact-valued and upper hemicontinuous, where \(X\subseteq\mathbb R^n\) and \(Y\subseteq\mathbb R^m\). If \(S\subseteq X\) is compact, then \[\varphi(S):=\bigcup_{x\in S}\varphi(x)\] is compact in \(Y\).
Blueprint. Start with an arbitrary sequence in \(\varphi(S)\) and choose its input witnesses in \(S\). Compactness supplies a convergent input subsequence. Upper hemicontinuity confines the corresponding output values to a bounded neighborhood of the limiting value set; extract a convergent output subsequence, then use the closed-graph lemma to retain its limit in the image.
The theorem needs all three elements: compact \(S\) supplies a convergent input subsequence, uhc controls nearby values, and compact-valuedness supplies both the local boundedness and the separation used by the closed-graph lemma. The conclusion concerns the union \(\varphi(S)\), not a single value \(\varphi(x)\).
Assumption audit and proof blueprint
- Closed plus bounded characterizes compactness only in \(\mathbb R^n\).
- Bolzano–Weierstrass starts from boundedness; closedness is what retains the limit in a set.
- The Intermediate Value Theorem needs a continuous real-valued function on a closed interval and a target \(y\) between the endpoint values.
- Continuity, usc/lsc, and uhc/lhc concern different objects. The first two concern scalar or vector function values; hemicontinuity concerns set values.
- A single \(\delta\) for the entire domain is uniform continuity, not ordinary pointwise continuity.
Blueprint: prove compactness of a set or image.
- Name the ambient Euclidean space.
- For a set, prove closedness and boundedness, or use the sequential test.
- For a correspondence image, start with an arbitrary output sequence and choose its input witnesses.
- State exactly where compactness, continuity, or hemicontinuity enters.
- Conclude that the extracted limit lies in the required set.
Exit tickets
- Give one sequence that proves \((0,1)\) is not closed in \(\mathbb R\).
- State the difference between usc of \(f\) and uhc of \(\varphi\) in one sentence.
- Why does compactness of the domain matter for turning continuity into uniform continuity?
- \(x_k=1/(k+1)\in(0,1)\) converges to \(0\notin(0,1)\). 2. Usc controls upward jumps of numerical values; uhc prevents nearby value sets from acquiring points outside any open neighborhood of the limiting value set. 3. It lets one extract a convergent subsequence from a purported failure of uniform control, which continuity then contradicts.