A basic knowledge of topology is very useful in the study of differential geometry - it provides a language and structure for notions relating to continuity and allows us to generalise from Euclidean submanifolds to manifolds as objects in their own right. Our focus is in what is often called point set topology. We won't be so interested in algebraic topology, and in particular we won't spend much time discussing coffee cups and donuts.
Here we review some of the most relevant topics. Some topics may not be covered in standard courses, and the perspective taken as well as some of the notation may also be unfamiliar. Thus even if you have a solid background in linear algebra it's worth at least taking a look. Throughout the course if you encounter some linear algebra you're not comfortable with, you can always return here to see if it's explained here.
The distance function (or metric ) on \(\mathbb{R}^n\) is the function \(d : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}\): \[ d(x, y) = \|x - y\| = \sqrt{(x_1 -y_1)^2 + \cdots + (x_n - y_n)^2}. \]
Let \(x \in \mathbb{R}^n\) and let \(r > 0\). The open ball \(\mathbb{B}_r(x)\)of radius \(r\) and centre \(x\) is the set of points \(y \in \mathbb{R}^n\) strictly within distance \(r\) of \(x\). That is \[ \mathbb{B}_r(x) = \{y \in \mathbb{R}^n : \|x - y\| < r\}. \]
Prove the triangle inequality for \(d\). That is, for all \(x, y, z \in \mathbb{R}^n\), \[ d(x, y) \leq d(x, z) + d(z, y). \]
Show that for any \(y \in \mathbb{B}_r(x)\) there exists an \(s > 0\) such that \(\mathbb{B}_s(y) \subseteq \mathbb{B}_r(x)\). Hint : use the triangle inequality.
Sometimes it's more convenient to work with (hyper) cubes than with balls.
The (hyper) cube \(\mathbb{C}_r(p)\) of width \(r > 0\) and centre \(p\) is the set \[ \mathbb{C}_r(p) = \{q : \max_{i=1,\dots n} |q_i - p_i| < r\}. \]
Let \(n = 2\). Then \[ \mathbb{C}_r ((0, 0)) = (-r, r) \times (-r, r) = \{(x,y) : -r < x < r, -r < y < r\} \]
Here \(p = (p_1, p_2) = (0, 0)\) and \(q = (q_1, q_2) = (x, y)\). Observe that \(\max_{i=1,2} |q_i| < r\) if and only if \(|x|, |y| < r\) if and only if \((x, y) \in (-r, r) \times (-r, r)\).
Show that \[ \mathbb{C}_r(p) = (p_1 - r, p_1 + r) \times \cdots \times (p_n - r, p_n + r). \]
For any \(r > 0\) and any \(p \in \mathbb{R}^n\) we have \(\mathbb{B}_r(p) \subseteq \mathbb{C}_r(p)\).
Let \(q \in \mathbb{B}_r(p)\) so that \((q_1 - p_1)^2 + \dots + (q_n - p_n)^2 = \|q - p\|^2 < r^2\). Then for each \(i\) we have \[ (q_i - p_i)^2 < r^2 \Rightarrow |q_i - p_i| < r. \]
Taking the maximum over \(i\) then gives \[ \max_{i=1,\dots, n} |q_i - p_i| < r \] hence \(q \in \mathbb{C}_r(p)\).
A \(U \subseteq \mathbb{R}^n\) is open if for every \(p \in U\) there is a ball \(\mathbb{B}_r(p)\) centred on \(p\) with radius \(r > 0\) such that \(\mathbb{B}_r(p) \subseteq U\).
The exercise of the previous section shows that open balls are open.
Let \(U = \{(x, y) \in \mathbb{R}^2 : y > x\}\). Then \(U\) is open.
To see this, let \(p_0 = (x_0, y_0) \in U\) and let \(r = \tfrac{y_0 - x_0}{2} > 0\).
Let \(p = (x, y) \in \mathbb{C}_r(p_0)\) so that \(-r < x-x_0, y-y_0 < r\). Then
\begin{align*} y - x &= (y-y_0) - (x-x_0) + y_0 - x_0 \\ &= (y-y_0) - (x-x_0) + 2r \\ &> -r - r + 2r = 0 \end{align*}hence \(p \in U\). Since this is true for any \(p \in \mathbb{C}_r(p_0)\) we obtain, \[ \mathbb{B}_r(p_0) \subseteq \mathbb{C}_r(p_0) \subseteq U \] and \(U\) is open.
Show that the following sets are open:
Show that a set \(U\) is open if and only if for every \(p \in U\) there exists an \(r > 0\) such that \[ \mathbb{C}_r(p) \subseteq U. \]
Hint : One implication follows from \(\mathbb{B}_r(p) \subseteq \mathbb{C}_r(p)\). To obtain the reverse implication, show that \(\mathbb{C}_r(p) \subseteq \mathbb{B}_{nr}(p)\).
Show that a set \(U\) is open if and only if \(U\) may be written as a union of open balls. That is, there is a set \(\{\mathbb{B}_{\alpha}\}_{\alpha \in A}\) of open balls such that \[ U = \bigcup_{\alpha \in A} \mathbb{B}_{\alpha}. \]
Hint : You don't need to be efficient with the choice of index set \(A\)!
Show that we can replace balls by cubes.
A set \(C \subseteq \mathbb{R}^n\) is closed if it is the complement of an open set. Equivalently, \(C\) is closed if it's complement \(C^c = \mathbb{R}^n \backslash C\) is open.
Show that the closed ball , \[ \overline{\mathbb{B}_r(x)} = \{y \in \mathbb{R}^n : \|y - x\| \leq r\} \] is closed.
In the case \(n = 1\), show that the open ball, \(\mathbb{B}_r(x)\) is the open interval \((x-r, x+r)\). Show that \(\overline{\mathbb{B}_r(x)}\) is the closed interval \([x-r, x+r]\).
Let \(U \subseteq \mathbb{R}^n\) be an open set and let \(f : \mathbb{R}^n \to \mathbb{R}^k\) be any function. Let \(x_0 \in U\) and let \(L \in \mathbb{R}^k\).
We say the limit of \(f\) as \(x\) approaches \(x_0\) equals \(L\) if for every open set \(W \subseteq \mathbb{R}^k\) containing \(L\), there is an open set \(V \subseteq U\) containing \(x_0\) such that \(f(V \backslash \{x_0\}) \subseteq W\). In this case we write \[ \lim_{x\to x_0} f(x) = L. \]
Otherwise, if there is not such \(L\), we say the limit does not exist.
The idea here is that we can zoom in around \(L\) be choosing a small open set \(W\) containing \(L\) - e.g. a ball of radius \(\epsilon\) for a small \(\epsilon\). Then no matter how closely we zoom in around \(L\), we can zoom in close enough to \(x_0\) (i.e. by restricting to the open set \(V\)) so that \(f\) all the points in \(V\) (except possibly \(x_0\) itself) are mapped into \(W\).
The whole point is that for every set \(W\), there exists a suitable set \(V\).
The reason we exclude \(x_0\) in the conclusion \(f(V \backslash \{x_0\}) \subseteq W\) is that the value \(f(x_0)\) is irrelevant to the limit; we only care about what happens as \(x\) approaches \(x_0\), but not what happens at \(x_0\).
The next example works through a relatively simple limit calculation. It's written out with a lot of detail and explanation about what's going on. After the example, there is a theorem that makes everything much easier and then the example is done again much more simply by using the theorem. If you find this example a bit tedious to work through, then feel free to skip it. I won't be offended - to be honest it was a bit tedious to write out!
Let \(f : \mathbb{R} \to \mathbb{R}\) be defined by \(f(x) = x^2\). Then \[ \lim_{x\to 2} f(x) = 4. \]
To see this, let \(W\) be an open set containing \(4\). We need to show that there exists an open set \(V\) containing \(2\) such that \(f(V) \subseteq W\). Since \(W\) is open, there exists an \(\epsilon > 0\) such that \((4-\epsilon, 4+\epsilon) = \mathbb{B}_{\epsilon} (4) \subseteq W\). For our open set \(V\) we will determine a positive real number \(\delta > 0\) such that \[ f(\mathbb{B}_{\delta} (2) \backslash \{2\}) \subseteq \mathbb{B}_{\epsilon} (4). \] That is we will take \(V = \mathbb{B}_{\delta}(2) = (2-\delta, 2+\delta)\). In fact given, \(\epsilon > 0\) we will choose \(\delta = \min\{1, \epsilon/6\}\).
To see why we make this choice, and that it does indeed suffice, first observe that since \(f(x) = x^2\), \(f(x) \in \mathbb{B}_{\epsilon}(4)\) if and only if \(|x^2 - 4| < \epsilon\) and \(x \in \mathbb{B}_{\delta}(2)\) if and only if \(|x-2| < \delta\). Thus for any \(x \in \mathbb{B}_{\delta}(2)\), we have \[ \lvert x^2 - 4| = |x-2||x+2| < \delta|x+2|. \]
We want the right hand side to be smaller than \(\epsilon\). Our choice \(\delta = \min\{1, \epsilon/6\}\) implies in particular that \(\delta \leq 1\). Thus for \(|x-2| < \delta \leq 1\) we have that \(1 < x < 3\) and hence \(|x+2| < 5\). Thus \[ \lvert x^2 - 4| = |x-2||x+2| < 5\delta \] whenever \(|x-2| < \delta\).
To finish we just require that in addition to \(\delta \leq 1\), we require that \(5 \delta < \epsilon\). Choosing \(\delta = \epsilon/6\) ensures this is true. Thus our choice \(\delta = \min\{1, \epsilon/6\}\) gives both \(\delta \leq 1\) and \(5\delta < \epsilon\) and hence that \[ f(V) = f(\mathbb{B}_{\delta}(2)) \subseteq \mathbb{B}_{\epsilon}(4) \subseteq W. \]
Therefore, for any open set \(W\) containing \(4\), there is an open set \(V\) (namely \(\mathbb{B}_{\delta}(2)\)) containing \(2\) such that \(f(V \backslash \{2\}) \subseteq W\). Thus by the definition of limits, we obtain \[ \lim_{x\to 2} x^2 = 4. \]
The approach in the example leads us to an equivalent (and somewhat more practical) definition of limit.
Show that it in the definition of limits, we may replace "open set" by "open ball" to obtain an equivalent definition. That is, show \(\lim_{x\to x_0} f(x) = L\) if and only if
For every \(\epsilon > 0\), there exists a \(\delta > 0\) such that \[ \| x - x_0\| < \delta \Rightarrow \|f(x) - L\| < \epsilon. \] Put another way, for every open ball \(\mathbb{B}_{\epsilon}(L)\), there exists an open ball \(\mathbb{B}_{\delta} (x_0)\) such that \(f(\mathbb{B}_{\delta}(x_0)) \subseteq \mathbb{B}_{\epsilon}(L)\).
The example above was quite simple, but it was somewhat painful to work through the details. What we need is a theorem that makes life easier. This is furnished by:
Let \(a, b \in \mathbb{R}\) and let \(L, K \in \mathbb{R}^k\). Let \(U \subseteq \mathbb{R}^n\) be an open set and let \(x_0 \in U\). Let \(f, g : U \subseteq \mathbb{R}^n \to \mathbb{R}^k\) be functions such that \[ \lim_{x\to x_0} f(x) = L, \quad \lim_{x\to x_0} g(x) = K. \]
Then
We can now greatly simplify the example \(\lim_{x\to 2} x^2 = 4\). Let \(f(x) = x\) and let \(g(x) = x\). Then we have \[ \lim_{x\to 2} f(x) = \lim_{x\to 2} g(x) = 2 \] since for every \(\epsilon > 0\), letting \(\delta = \epsilon\) gives \[ \lvert x - 2| < \delta \Rightarrow |f(x) - 2| = |x - 2| < \delta = \epsilon. \] Likewise for \(g\).
Then by the theorem, \[ \lim_{x\to 2} x^2 = \lim_{x\to 2} f(x) g(x) = 2 \cdot 2 = 4. \]
Use the theorem to show that if \(f(x) = a_nx^2 + \cdots + a_1 x + a_0\) is any polynomial, then for any \(x_0 \in \mathbb{R}\), \[ \lim_{x\to x_0} f(x) = f(x_0). \]
Show that \(\lim_{x \to x_0} f(x) = L\) if and only if \(\lim_{x\to x_0} |f(x) - L| = 0\).
Let
\begin{equation*} f(x) = \begin{cases} 0, & x < 0 \\ 1, & x \geq 0 \end{cases} \end{equation*}Show that \(lim_{x\to 0} f(x)\) does not exist. That is for every \(L \in \mathbb{R}\), it is not true that \(\lim_{x\to 0} f(x) = L\).
Let \(f(x) = \sin(1/x)\). Show that \(\lim_{x\to 0} f(x)\) does not exist.
The topology of \(\mathbb{R}^n\) is the collection of all open sets.
By definition all of \(\mathbb{R}^n\) is open hence is part of the topology. Also by definition the empty set \(\emptyset\) is part of the topology. The empty set is vaccuously open since it has no elements and hence it is true that \(\mathbb{B}_r(x)\) is contained in the empty set for every \(x \in \phi\) and every \(r > 0\). If you feel uncomfortable with this claim, one way to accept it is to simply say that we define the empty set to be open!
The topology of \(\mathbb{R}^n\) is closed under arbitrary unions and finite intersections. That is, for any collection \(\{U_{\alpha}\}_{\alpha \in A}\) of open sets, \[ \bigcup_{\alpha \in A} U_{\alpha} \] is open, and for any finite collection \(\{U_i\}_{i=1}^k\), \[ \bigcap_{i=1}^k U_i \] is open.
Equivalently, the collection of closed sets is closed under finite unions and arbitrary intersections.
Let \(U = \bigcup_{\alpha \in A} U_{\alpha}\) and let \(x \in U\) be an arbitrary point. Then \(x \in U_{\beta}\) for some \(\beta \in A\). Since \(U_{\beta}\) is open, there is an \(r > 0\) such that \(\mathbb{B}_r(x) \subseteq U_{\beta}\). But \(U_{\beta} \subseteq U\), hence \[ \mathbb{B}_r(x) \subseteq U_{\beta} \subseteq U \] and thus \(U\) is open.
Now consider a finite collection \(U_i\) and let \(U = \bigcap_{i=1}^k U_i\) If \(U = \emptyset\), then \(U\) is open by definition. Otherwise, let \(x \in U\). Then \(x \in U_i\) for every \(i\). Since each \(U_i\) is open, for each \(i\) we get an \(r_i > 0\) such that \(\mathbb{B}_{r_i}(x) \subseteq U_i\). Since there are only finitely many \(i\), \(r = \min \{r_i : i = 1, \dots, n\}\) is positive. Then for each \(i\), \[ \mathbb{B}_r (x) \subseteq \mathbb{B}_{r_i} (x) \subseteq U \] and hence \(U\) is open.
Consider the collection of open intervals \(U_i = \big(-\tfrac{1}{i}, \tfrac{1}{i}\big) \subseteq \mathbb{R}\) for \(i = 1, 2, \dots, \). Show that \(\bigcap_{i=1}^{\infty} U_i\) is not open. Thus an intersection over an infinite collection may not be open. This doesn't mean an infinite intersection can't be open - just that not every infinite intersection is open. Give an example of an infinite collection where the intersection is open.
A frequently used concept in topology is that of an open cover.
Let \(S \subseteq \mathbb{R}^n\) be any subset. An open cover of \(S\) is a collection of open sets \(\{U_{\alpha}\}_{\alpha \in A}\) such that \[ S \subseteq \bigcup_{\alpha \in A} U_{\alpha}. \]
Let \(U \subseteq \mathbb{R}^n\) be an open set. Then by definition, for every \(p \in U\) there exists an \(r > 0\) such that \(\mathbb{B}_r (p) \subseteq U\). Then an open cover for \(U\) is \(\{\mathbb{B}_r (p) : p \in U\}\)!
Let's have a rapid fire introduction to/review of topological spaces.
A topological space is a set \(X\) equipped with a collection, \(\mathcal{O}\) of subsets of \(X\) such that
The collection \(\mathcal{O}\) is called a topology , and the sets \(U \in \mathcal{O}\) are called open sets .
In words, the requirements are that the empty set and the whole space are open, arbitrary unions of open sets are open, and finite intersections of open sets are open. A topology is an axiomatisation of the notion of open sets of Euclidean space from the previous section. But there are many and varied topological spaces throughout mathematics, some of which have very different behaviour to Euclidean topologies.
Let \(X\) be a topological space and let \(Y \subseteq X\). The induced topology on \(Y\) is the collection of sets \[ \mathcal{O}_Y = \{U \cap Y : U \in \mathcal{O}_X\} \] We call the pair \((Y, \mathcal{O}_Y)\) a (topological) subspace.
Show that \(\mathcal{O}_Y\) is a topology.
The axioms of topology allow for certain behaviour that is not appropriate for developing calculus. Recall that derivatives are defined using the linear structure of Euclidean space via \[ D_V f (p) = \lim_{h \to 0} \frac{f(p + hV) - f(p)}{h}. \] We will need some sort of underlying Euclidean structure on our space in order to generalise this definition. From this perspective, it's natural to consider topological spaces that share some of the properties of Euclidean space.
A Hausdorff topological space is a topological space \(X\) such that for any \(p, q \in X\) \(p \neq q\), there exists disjoint open neighbourhoods \(U, V\) of \(p, q\) respectively. That is there exist open sets \(U, V\) with \(p \in U\) and \(q \in V\) such that \(U \cap P = \emptyset\).
In Euclidean space, such neighbourhoods are furnished by \(U = \mathbb{B}_{r/4}(p)\) and \(V = \mathbb{B}_{r/4}(q)\) where \(r = d(p, q) = \|p - q\|\). For subspaces of \(\mathbb{R}^n\), i.e. subsets \(S \subseteq \mathbb{R}^n\) equipped with the induced topology, we just intersect these neighbourhoods with \(M\): \(U = \mathbb{B}_{r/4}(p) \cap S\) and \(V = \mathbb{B}_{r/4}(q) \cap S\). In this way, both Euclidean spaces and subspaces of Euclidean space are Hausdorff spaces.
The Hausdorff condition is natural then to carry over to generalisations of Euclidean space, inheriting desirable properties such as uniqueness of limits. The Hausdorff condition is not automatic for topological spaces - there are topological spaces that are not Hausdorff. Though such spaces are not relevant to us, observe their existence forces us to impose the extra Hausdorff condition on our topological spaces.
The next condition is a rather more technical condition that allows for certain constructions (e.g. partitions of unity). Although the topologies we consider (e.g. on Euclidean space) have uncountably infinitely many open sets, all such sets may be generated by only countably many sets. This has certain technical advantages, in particular ensuring that partitions of unity exist.
We won't dwell too much on this notion. If you haven't studied topological spaces before, don't worry too much about it for now. Just treat it as some extra, necessary technical condition that will help us out later. For the record, we give the necessary technicalities. Feel free to skip them for now.
First, let us define the notion of a base for a topology.
A base \(\mathcal{B}\) for a topological space \((X, \mathcal{O})\) is a subset \(\mathcal{B} \subseteq \mathcal{O}\) of sets such that every \(U \in \mathcal{O}\) may be written (in general non-uniquely) as a union of elements of \(\mathcal{B}\): \[ U = \bigcup_{\alpha} U_{\alpha} \] where \(U_{\alpha} \in \mathcal{B}\) for each \(\alpha\).
One base for the topology on \(\mathbb{R}^n\) is the set of open balls \[ \mathcal{B}_1 = \{(\mathbb{B}_r(x) : x \in \mathbb{R}^n, r \in \mathbb{R}, r > 0\}. \] Another base is the set of open hypercubes, \[ \mathcal{B}_2 = \{(\mathbb{C}_r(x) : x \in \mathbb{R}^n, r \in \mathbb{R}, r > 0\} \] where \(\mathbb{C}_r(x)\) denotes the hypercube of width \(r/2\) centred on \(x\).
A second countable topological space is a topological set with a countable base.
Euclidean space and it's subspaces are second countable topological spaces.
A base is \[ \mathcal{B} = \{\mathbb{B}_r(x) : x \in \mathbb{Q}^n, r \in \mathbb{Q}, r > 0\}. \] That is, the set of balls centred on rational points and with rational radius.
If you know about countable sets and the fact that \(\mathbb{Q}\) is dense in \(\mathbb{R}\), prove the base given above is indeed a base.
Let \(U \subseteq \mathbb{R}^n\) be an open set and let \(p \in U\). A function \(f : U \to \mathbb{R}^k\) is continuous at \(p\) if \[ \lim_{q \to p} f(q) = f(p). \]
Here we assert that the limit exists and is equal to the function value at \(p\).
Show that any polynomial \(f(p) = a_np^n + \cdots + a_1 p + a_0\) is continuous at every \(p \in \mathbb{R}^n\).
A function \(f(p) = (f_1(p), \dots, f_k(p)\) from \(\mathbb{R}^n\) to \(\mathbb{R}^k\) is continuous at \(p\) if and only if each of the component functions \(f_i\), \(i=1,\dots,k\) are continuous at \(p\).
We have that \[ \|f(q) - f(p)\|^2 = (f_1(q) - p_1)^2 + \cdots + (f_k(q) + p_k)^2. \] Thus \(\lim_{q\to p} f(q) = f(p)\) if and only if \(\lim_{q\to p} \|f(q) - f(p)\| = 0\) if and only if for every \(i = 1, \dots, k\), \(\lim_{q\to p}\|f_i(q) - f_i(p)\| = 0\) if and only if for every \(i = 1, \dots, k\), \(\lim_{q\to p} f_i(q) = f(p)\).
Let \(U \subseteq \mathbb{R}^n\) be an open set. A function \(f : U \to \mathbb{R}^k\) is continuous if it is continuous at every point \(p \in U\).
A function \(f : U \to \mathbb{R}^k\) is continuous if and only if for every open set \(V \subseteq \mathbb{R}^k\), the pre-image, \[ f^{\ast} (V) := \{q \in U : f(q) \in V\} \] is open.
Suppose \(f\) is continuous and let \(V \subseteq \mathbb{R}^k\) be an open set. We need to show that \(f^{\ast}(V) \subseteq \mathbb{R}^n\) is open. By definition, this means that for every \(p \in f^{\ast}(V)\) there exists an \(r > 0\) such that \(\mathbb{B}_r(p) \subseteq V\).
Let \(p \in f^{\ast}(V)\) be arbitrary. Since \(f\) is continuous at \(p\), \(\lim_{q\to p} f(q) = f(p)\). By the definition of limits, for any open set \(Z \subseteq \mathbb{R}^k\) with \(f(p) \in Z\), there is an open set \(W\) with \(p \in W\) such that \(f(W \backslash \{p\}) \subseteq Z\). Applying this to \(Z = V\), there is an open set \(W \subseteq U\) with \(p \in W\) and such that \(f(W \backslash \{p\}) \subseteq V\). Thus \(W \backslash \{p\} \subseteq f^{\ast}(V)\). Since \(p \in f^{\ast} (V)\), in fact we have that \(W \subseteq f^{\ast} (V)\).
But now, since \(W\) is open there exists \(r > 0\) with \(\mathbb{B}_r(p) \subseteq W \subseteq f^{\ast}(V)\). Since \(p \in f^{\ast} (V)\) was arbitrary, we see that \(f^{\ast} (V)\) is open.
Conversely, suppose \(f^{\ast} (V)\) is open for every open \(V \subseteq \mathbb{R}^k\). Let \(p \in U\). To show that \(f\) is continuous at \(p\) we need to show that \(\lim_{q\to p} f(q) = f(p)\). That is, we need to show that for every open set \(V \in \mathbb{R}^k\) containing \(f(p)\), there is an open set \(W \subseteq U\) containing \(p\) and such that \(f(W \backslash \{p\}) \subseteq V\). But this is immediate since by hypothesis the set \(W = f^{\ast} (V)\) is open, hence \(f(W \backslash \{p\}) \subseteq W = V \subseteq V\).
Generalising continuous functions from Euclidean space to topological spaces is straightforward.
Let \(X, Y\) be topological spaces and let \(f : X \to Y\). We say \(f\) is continuous if \(f^{\ast}(U) \subseteq X\) is open in \(X\) for every subset \(U \subseteq Y\) that is open in \(Y\).
A set \(S \subseteq \mathbb{R}^n\) is bounded if there exists an \(R > 0\) such that \(C \subseteq \mathbb{B}_R(0)\). That is, for every \(p \in C\), \(\|p\| \leq R\).
Bounded sets play an important role in Euclidean topology through compactness. Here are three different definitions of compactness.
A set \(K \subseteq \mathbb{R}^n\) is Euclidean compact if it is closed and bounded.
A set \(K \subseteq \mathbb{R}^n\) is sequentially compact if for every sequence of points \((p_n)_{n \in \mathbb{N}}\), there exists a convergent subsequence \(p_{n_j}\).
A set \(K \subseteq \mathbb{R}^n\) is Topologically compact if for every open cover \(\{U_{\alpha}\}_{\alpha \in A}\), there is a finite subcover \(\{U_{\alpha_1}, \dots, U_{\alpha_k}\}\). That is whenever \(K \subseteq \bigcup_{\alpha \in A\} U_{\alpha}\), there exists finitely many \(\alpha_1, \dots, \alpha_k \in A\) such that \(K \subseteq \bigcup_{i=1}^k U_{\alpha_i}\).
Let \(K \subseteq \mathbb{R}^n\). Then \(K\) is Euclidean compact if and only if it is sequentially compact if and only if it is topologically compact.
There are other situations such as general metric and topological spaces where the three definitions are not equivalent. In a general Hausdorff topological space (of which Euclidean space and metric spaces are special cases), topological compactness implies sequential compactness implies Euclidean compactness. However, the reverse implications are false in general.
The open ball \(\mathbb{B}_r(x)\) is not closed, hence not compact. The closed ball \(\overline{\mathbb{B}_r(x)}\) is closed and bounded, hence compact.
The set \(\{y = x^2, -1 \leq x \leq 1\} \subseteq \mathbb{R}^2\) is closed and bounded, hence compact. The set \(\{y = x^2\} \subseteq \mathbb{R}^2\) is closed, but not bounded, hence is not compact.
The most important results here for us however are to do with continuity.
Let \(U \subseteq \mathbb{R}^n\) be an open set and let \(f : U \to \mathbb{R}^k\) be continuous. Let \(K \subseteq \mathbb{R}^n\) be a compact set with \(K \subseteq U\).
Then
One important and very useful construction is that of bump functions and cut off functions, For continuous functions, this is relatively straightforward. Later we will construct a smooth version which is a little more technical.
Let \(f : \mathbb{R}^n \to \mathbb{R}\) be a continuous function. The support, \(\operatorname{supp} f\) of \(f\) is the closure of the set point on which \(f\) is non-zero: \[ \operatorname{supp} (f) = \overline\{p \in \mathbb{R}^n : f(p) \neq 0\}. \]
Let \(U \subseteq \mathbb{R}^n\) be an open set. A continuous cutoff function supported in \(U\) is a continuous function \(\rho : \mathbb{R}^n \to \mathbb{R}\) such that
Thus outside of \(U\), \(\rho\) equals \(0\) so it is "cut off" outside \(U\).
Often we also require that \(\rho \equiv 1\) on some open subset \(V \subseteq U\) with \(\overline{V} \subseteq U\). In that case we might think of \(\rho\) as a continuous approximation of the indicator function \(\chi_V\) which is equal to \(1\) on \(V\) and \(0\) everywhere else. This latter is not continuous, and \(\rho\) serves as a continuous approximation - especially if \(U \backslash V\) is "small" in some sense.
Let \(U \subseteq \mathbb{R}^n\) be an open set. Then there exists a continuous cutoff function support in \(U\).
Pick any \(p \in U\) and \(r > 0\) such that \(\mathbb{B}_r(p) \subseteq U\). Then a suitable \(\rho\) is given by
\begin{equation*} \rho(q) = \begin{cases} \left(1 - \frac{\|q - p\|}{r}\right), & q \in \mathbb{B}_r(p) \\ 0, & q \notin \mathbb{B}_r (p). \end{cases} \end{equation*}We can similarly construct a continuous cut off function that is supported on a ball, and is identically \(1\) on a smaller ball. For example,
\begin{equation*} \rho(q) = \begin{cases} 1, & q \in \mathbb{B}_{1/2} (p) \\ 2(1 - \|q - p\|), & q \in \overline{\mathbb{B}_1(p) \backslash \mathbb{B}_{r/2}(p)} \\ 0, & q \notin \mathbb{B}_1(p). \end{cases} \end{equation*}Then \(\operatorname{supp} (\rho) \subseteq \mathbb{B}_1(p)\) \(\rho \equiv 1\) on \(\mathbb{B}_{1/2}(p)\).
Let \(p \in \mathbb{R}^n\) and let \(0 < r_1 < r_2\). Construct a continuous cutoff function \(\rho\) such that
Let \(U, V \subseteq \mathbb{R}^n\) be open sets such that \(\overline{V} \subseteq U\). Then there exists a continuous cut off function \(\rho\) such that
We won't prove the theorem here. For reference, it is known as Urysohn's lemma.
Let \(C = \{U_{\alpha}\}_{\alpha}\) be a collection of open sets in \(\mathbb{R}^n\). A continuous partition of unity (abbreviated p.o.u.\) subordinate to \(C\) is a collection of continuous cut off functions \(\{\rho_{\alpha} : U_{\alpha} \to \mathbb{R}\}_{\alpha \in A\}\) such that for every \(p \in \bigcup_{\alpha \in A} U_{\alpha}\),
Let \(C = \{U_{i}\}_{i=1,\dots,k\}\) be a finite collection of open sets. Then there exists a partition of unity subordinate to \(C\).
TBA.
Let \(K \subseteq \mathbb{R}^n\) be a compact set and let \(C = \{U_{\alpha}\}_{\alpha \in A}\) be an open cover. Then there exists a partition of unity subordinate to \(C\).
We won't prove the corollary here. It's almost immediate since \(K\) compact means there is a finite subcover. But this will only give continuous cut off function \(\rho_{\alpha_i}\), \(i = 1, \dots, k\) for a finite subset of \(A\). To be a true partition of unity, we need to construct \(\rho_{\alpha}\) for every \(\alpha \in A\). This takes a little more work, but is not too involved. We omit the details here though.
It turns out that partitions of unity exist for any open cover.
Let \(C = \{U_{\alpha}\}_{\alpha \in A\}\) be an arbitrary collection of open sets. Then there exists a partition of unity subordinate to \(C\).
The proof is again omitted. The starting point is Urysohn's lemma, but some extra work is now required compared to the compact case since we can't guarantee a finite subcover.