MAT257Y1/Unit 1
Unit 1 · Analysis II

The Topology of $\mathbb{R}^n$

MAT158H1 built analysis on the order structure of $\mathbb{R}$: every argument leaned on $<$. In $\mathbb{R}^n$ there is no order. What survives is distance, and this unit rebuilds the vocabulary of analysis — open, closed, limit point, closure — from distance alone.

By the end you can
Why it matters & where it comes from
Why it exists

Every theorem in the rest of the year quantifies over neighbourhoods. Without a precise notion of "near", the inverse function theorem cannot even be stated, since it asserts something on a neighbourhood whose size nobody specifies.

In the real world

Optimisation algorithms terminate on a tolerance, which is a statement about open balls. Numerical stability is continuity made quantitative. Machine learning generalisation bounds are covering arguments over compact parameter sets.

Story & open questions

Hausdorff axiomatised topology in 1914 by taking neighbourhoods as primitive and forgetting distance entirely. Everything in this unit survives that abstraction; almost nothing in the next unit does, which is why general topology (MAT327H1) has to work so much harder for compactness.

Norms and the metric

Definition

For $x\in\mathbb{R}^n$ the Euclidean norm is $\|x\|=\sqrt{\sum_{i=1}^nx_i^2}$, and the distance between $x$ and $y$ is $\|x-y\|$. The open ball of radius $r$ about $a$ is $B_r(a)=\{x:\|x-a\|<r\}$.

Theorem Equivalence of norms

Write $\|x\|_1=\sum_i|x_i|$ and $\|x\|_\infty=\max_i|x_i|$. Then

$\|x\|_\infty\le\|x\|\le\|x\|_1\le n\|x\|_\infty$.

So the three norms have the same open sets, the same convergent sequences and the same continuous functions. Any of them may be used wherever it is convenient.

For the first: $\|x\|^2=\sum_ix_i^2\ge\max_ix_i^2$. For the second: $\|x\|^2=\sum_ix_i^2\le\left(\sum_i|x_i|\right)^2$ since the cross terms are non-negative. For the third: each $|x_i|\le\|x\|_\infty$ and there are $n$ terms. ∎
Why the balls look different but behave the same. The $\|\cdot\|_\infty$ ball is a cube and the $\|\cdot\|_1$ ball is a diamond. The theorem says each fits inside a scaled copy of the other, so a statement of the form "there exists a ball contained in $U$" does not care which norm you used. In infinite dimensions this fails, and it is one of the main reasons functional analysis is harder.

Open and closed sets

Definition

$U\subseteq\mathbb{R}^n$ is open if for every $a\in U$ there is $r>0$ with $B_r(a)\subseteq U$. A set is closed if its complement is open.

Theorem What survives unions and intersections
Arbitrary unions of open sets are open; finite intersections of open sets are open.
Arbitrary intersections of closed sets are closed; finite unions of closed sets are closed.

For a union, a point lies in some member, which already supplies a ball. For a finite intersection, take the smallest of finitely many radii — which is why finiteness is needed, since an infinite family can have radii shrinking to $0$.

Example Infinite intersection fails

$\bigcap_{n=1}^\infty\left(-\tfrac1n,\tfrac1n\right)=\{0\}$, an intersection of open intervals that is not open. Every candidate radius is killed by taking $n$ large enough.

Example Neither, and both
$[0,1)$ in $\mathbb{R}$ is neither open nor closed: no ball about $0$ stays inside, and $1$ is a limit point not in the set.
$\emptyset$ and $\mathbb{R}^n$ are both open and closed. In $\mathbb{R}^n$ these are the only such sets, which is exactly the statement that $\mathbb{R}^n$ is connected (Unit 3).

Limit points, closure and boundary

Definition

$a$ is a limit point of $S$ if every ball $B_r(a)$ contains a point of $S$ other than $a$. The closure $\overline{S}$ is $S$ together with its limit points; the interior $\operatorname{int}S$ is the set of points having a ball inside $S$; the boundary is $\partial S=\overline{S}\setminus\operatorname{int}S$.

Theorem Three descriptions of closed

For $S\subseteq\mathbb{R}^n$ the following are equivalent:

(1) $S$ is closed, that is $\mathbb{R}^n\setminus S$ is open.
(2) $S$ contains all its limit points, so $S=\overline{S}$.
(3) Whenever $x_k\in S$ and $x_k\to x$, the limit $x$ lies in $S$.
(1) ⇒ (3): if $x\notin S$ then some ball about $x$ misses $S$, so no sequence from $S$ can converge to $x$.
(3) ⇒ (2): a limit point $a$ has a point $x_k\in S\cap B_{1/k}(a)$ for each $k$, giving $x_k\to a$, so $a\in S$.
(2) ⇒ (1): if $x\notin S$ then $x$ is not a limit point, so some ball misses $S$ and lies in the complement. ∎
Example Compute all three

Let $S=\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}\cup\{(1,0)\}$.

$\operatorname{int}S$ is the open disc: the point $(1,0)$ has no ball inside $S$.
$\overline{S}$ is the closed disc, since every point of the unit circle is a limit of interior points.
$\partial S$ is the whole unit circle. Note $(1,0)$ belongs to $S$ and to $\partial S$ at once — a boundary point can be in or out of the set.
Sequences are the practical tool

Condition (3) is almost always the fastest way to prove a set is closed: take a convergent sequence in it and show the limit stays. Conditions (1) and (2) are better for the theory, but on a problem set reach for sequences.

Common pitfalls

• Treating "not open" as "closed". Most sets are neither, and $\mathbb{R}^n$ and $\emptyset$ are both.
• Confusing limit points with elements. A limit point need not belong to the set, and an isolated element of the set is not a limit point of it.
• Using finiteness where it fails. Infinite intersections of open sets can be closed.
• Forgetting that open and closed are relative. $[0,1)$ is not open in $\mathbb{R}$ but is open as a subset of $[0,\infty)$ with the subspace topology.

Practice

Original problems written for this guide in the style of Spivak and Munkres exercises. Write a full solution before opening the answer.

Practice 1 · Classify

Is $\{(x,y):xy>1\}$ open, closed, or neither?

Answer
Open. It is the preimage of the open set $(1,\infty)$ under the continuous map $(x,y)\mapsto xy$, and preimages of open sets under continuous maps are open (Unit 3).
Practice 2 · Closure

Find $\overline{S}$ for $S=\{1/n:n\in\mathbb{N}\}\subset\mathbb{R}$.

Answer
$S\cup\{0\}$. The only limit point is $0$, and no element of $S$ is a limit point of $S$ since each is isolated.
Practice 3 · Norm equivalence

Prove $\|x\|_\infty\le\|x\|\le\sqrt n\,\|x\|_\infty$.

Solution
Let $M=\|x\|_\infty$. Then $\|x\|^2=\sum_ix_i^2\ge M^2$ giving the left inequality, and $\|x\|^2\le nM^2$ giving the right after taking square roots.
Practice 4 · Boundary

Find $\partial\mathbb{Q}$ inside $\mathbb{R}$.

Solution
All of $\mathbb{R}$. Every interval contains both rationals and irrationals, so $\overline{\mathbb{Q}}=\mathbb{R}$ while $\operatorname{int}\mathbb{Q}=\emptyset$.
Practice 5 · Open balls are open

Prove $B_r(a)$ is open.

Solution
For $x\in B_r(a)$ put $s=r-\|x-a\|>0$. If $\|y-x\|<s$ then $\|y-a\|\le\|y-x\|+\|x-a\|<r$, so $B_s(x)\subseteq B_r(a)$. The triangle inequality is doing all the work.
Practice 6 · Closure of a ball

Is $\overline{B_r(a)}$ always the closed ball $\{x:\|x-a\|\le r\}$?

Solution
In $\mathbb{R}^n$ with a norm, yes: any point at distance exactly $r$ is a limit of points on the segment towards $a$. In a general metric space it can fail — in a discrete metric, $B_1(a)=\{a\}$ is already closed while the closed ball is everything.
Practice 7 · Interior of a union

Is $\operatorname{int}(A\cup B)=\operatorname{int}A\cup \operatorname{int}B$?

Solution
No, only $\supseteq$ holds. Take $A=\mathbb{Q}$ and $B=\mathbb{R}\setminus\mathbb{Q}$: both have empty interior, but their union is $\mathbb{R}$ whose interior is everything.
Practice 8 · Closed graph

Show the graph $\{(x,f(x)):x\in\mathbb{R}\}$ of a continuous $f:\mathbb{R}\to\mathbb{R}$ is closed in $\mathbb{R}^2$.

Solution
Take $(x_k,f(x_k))\to(x,y)$. Then $x_k\to x$, so $f(x_k)\to f(x)$ by continuity; limits are unique, so $y=f(x)$ and the limit is on the graph. Use the sequence criterion.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT257Y1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Use the current UofT MAT257Y1 calendar topics as the coverage spine, and pair lecture notes with Spivak's Calculus on Manifolds or Munkres' Analysis on Manifolds. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

Norms give balls, balls give open sets, complements give closed sets, and sequences give the working criterion for closedness. Unit 2 adds compactness on top of exactly this vocabulary.

Study routine. For every set you meet, name its interior, closure and boundary before doing anything else. Half the errors in this course come from applying a theorem to a set that fails its hypothesis.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Definition of open

State what it means for $U$ to be open.

Guide
Every point of $U$ has some open ball around it contained entirely in $U$.
Extension 2 · Where finiteness enters

Which operation on open sets requires finiteness, and why?

Guide
Intersection. You take the minimum of the radii, and an infinite family can have radii with infimum $0$.
Extension 3 · Sequence criterion

How do you show a set is closed using sequences?

Guide
Take any convergent sequence of points in the set and show its limit is also in the set.
Extension 4 · Consequence of equivalence

Why does norm equivalence matter for the rest of the course?

Guide
Because open sets, convergence and continuity are identical under all three norms, every proof may silently pick whichever norm is easiest — usually $\|\cdot\|_\infty$ for box arguments.
Back to IndexAll chapters Unit 02 →Compactness