Topology bits, a bunch of notes

This was originally going to be about another thing, but I needed to introduce (and learn) about topological notions beforehand. It became so long that it ended up being its own thing. It’s based on parts from two textbooks with a couple of my own proofs. May contain errors.

What’s a topology?

Topology

A topology on a set \(X\) is an arbitrary collection \(\tau\) of subsets of \(X\) that has the following three properties:

• The empty set and the set itself are contained in it

• The intersection of any two elements in \(\tau\) is also in it

• The union of any (finite or infinite) collection of sets in \(\tau\) is in \(\tau\)

(we are abusing notation and using \(\cup_{ A}\) for \(\cup_{a \in A} a\) )

We call \(U \subseteq X\) an open set if \(U \in \tau\) and we call it a closed set if \(X-U \in \tau\). Depending on the topology, a set might be both open and closed (take \(\varnothing\) for instance ) or it might not be either of them. We call a neighborhood of \(a\) any open set that includes \(a\).

The main intuition behind topologies and open sets is defining what it means when we say that a set includes the surroundings of any point in it. The 3 properties we demand for a topology follow from that motivation. We are not defining what the surroundings of a point really are, we are just building a way to discriminate the sets that have this property from the ones that don’t: this set has it, that one doesn’t, and so on.

Basis for Topology

Given a collection of sets \(\mathcal{B} \subseteq \mathcal{P}(X)\). We call it a basis of \(\tau\) if an arbitrary set is open iff it’s the union of elements in the basis. In other words \(\mathcal{B}\) is a basis of \(\tau\) if \(\tau\) is made of all the possible union of elements in \(\mathcal{B}\)

Using the concept of basis we can also talk about open sets in a more familiar language. Let \(\mathcal{B}\) be a basis of \(\tau\) then \(U\) is an open set iff each value \(x \in U\) is contained on some basis element \(B_x\) also in \(U\). That is \(x \in B_x \subseteq U\). This follows from the definition of basis.

Conditions for being a basis

Now, suppose that we are given an arbitrary collection of sets, and we want to consider the topology associated with that basis. Not any collection of sets is a valid basis for a topology. It might happen that the collection of all possible unions of elements in this collection is not a topology. Thus, we need to make sure that this collection obeys some properties in order to be considered a basis

\(\mathcal{B}\) is a basis for some topology on \(X\) iff:

• \(X = \bigcup_{ \mathcal{B}}\).

• \(B_1, B_2 \in \mathcal{B}\) then \(B_1 \cap B_2\) is the union of elements in \(\mathcal{B}\)

Basis for the same topology

Let \(\mathcal{B}_1\) be a basis for \(\tau_1\) on \(X\) and similarly \(\mathcal{B}_2\) a basis for \(\tau_2\). Then \(\tau_1 = \tau_2\) if only if \(\mathcal{B}_1 \subseteq \tau_2, \mathcal{B}_2 \subseteq \tau_1\). In other words, the elements of both bases are open in the other respective topology.

The forward direction is trivial. For the backward direction: \(\mathcal{B}_1 \subseteq \tau_2\) implies that any union of elements in \(\mathcal{B}_1\) is also in \(\tau_2\) (by definition of topology). Thus, \(\tau_1 \subseteq \tau_2\), same argument proves \(\tau_2 \subseteq \tau_1\)

Topology induced by a metric

Given an \(X\) and \(d\) a distance function \(d\), we can define a topology \(\tau\) leveraging this distance function:

\(U \subseteq X\) is open if for any \(x \in U\) there’s an \(\epsilon > 0\) such that the open ball \(B_{\epsilon}(x) = \{ y \mid d(x,y) < \epsilon \}\) is a subset of \(U\)

Intuitively speaking, a set is open if all points also have their surroundings included in it.

Equivalently, we can define this topology by defining (one of) its basis. The collection of open balls:

A useful property is that any element in an open ball (or union of open balls by extension) also has an open ball centered at it. That is, for any \(y \in B_{\epsilon}(x)\), there’s a \(B_{\epsilon^\prime}(y)\subseteq B_{\epsilon}(x)\)

Proof

Take \( y \in B_\epsilon (x) \), we know, by definition: $$d(x,y) = c < \epsilon $$ The trick is choosing a small enough neighborhood so that we can guarantee to remain in the set. Take \( \epsilon^\prime = \epsilon - c \) and \( B_{\epsilon^\prime }(y) \). It's easy to see that any element \(z\) on this ball is also contained in our original ball $$ d(y,z) < \epsilon^\prime \\ d(x,z) \leq d(x,y) + d(y,z) < c + \epsilon^\prime = \epsilon $$ So \( B_{ \epsilon^\prime }(y) \subseteq B_{\epsilon}(x) \)

Ok, let’s show that this is indeed a topology.

Note that all we are doing is constructing a generalization of the traditional notion of open sets in euclidean space given by \(X=R^k\) with the euclidean distance (or any other vector norm as they constitute different bases for the same topology).

A topological space that is induced by some distance function is called metrizable. On the other hand, if there is no such distance function then it is not metrizable. Curiously, there are topological spaces in this category. A trivial example is \(\tau = \{ X, \varnothing\}\) when \(X\) has more than one element.

This does not mean that each metric induces a different topology. One can have multiple metrics that lead to the same topology. Take for instance \(\|\|_2\) and \(\|\|_\infty\). We can bound:

Because of this, we have:

So if we have a ball centered at \(x\) in one metric we have another one in the other metric, thus they have the same open sets. (the same is true for any norm on \(R^n\) as they are all bound each other)

A small and silly warning on the term “open ball”: it’s not necessarily shaped like a ball at all! It’s just a term.

• An open interval is an open ball on the natural metric of the real line
• As we just saw, an open square is an open ball when using the infinity-norm as a metric
• The funny-looking half-open half-closed circle piece on the subspace topology example below is an open ball in the \(\tau_S\) topology.

Another cool point is that we don’t need to restrict ourselves to functions that follow \(d(x,y) = 0 \implies x=y\). Those spaces are called pseudometric spaces. \(\tau = \{ X, \varnothing\}\) is one example of a pseudometric space.

Some definitions around topologies

Discrete and trivial topology

The power set \(\mathcal{P}(X)\) is a topology on \(X\), here any set is considered open. This is called the discrete topology. Note that the set of isolated points forms a basis for this topology

On the other extreme \(\{ X, \varnothing\}\) is another topology. Here we are allowing only the whole set and the empty set to be open, anything else is not. This is called the trivial topology

Subspace topology

If you have a topology \(\tau\) on \(X\) you can focus on a subset \(S\) of \(X\) and talk about a topology restricted to that set. We called that a subspace topology

This leads to a new topology with new open sets. As an example, on the euclidean topology on the real line, \([0,1]\) is not an open set, but it is open in the subspace topology on \([0,1]\) Another similar example is given by the subspace topology on the closed unit circle. This subspace topology inherits the euclidean metric but some sets that were not open nor closed are now open.

Generally speaking, when talking about a subset of \(R^k\) we’ll assume we are using the topology taken the induced topology from euclidean space unless we specifically state otherwise

Finer and coarser topologies

Given two topologies \(\tau, \tau^\prime\) on \(X\), such that \(\tau \subseteq \tau^\prime\), we call \(\tau^\prime\) a finer topology than \(\tau\). Similarly, we call \(\tau\) a coarser topology than \(\tau ^\prime\). In short, a finer topology includes all the open sets of a coarser topology and then some.

Note that using these new words to refer to known concepts, two topologies are equal if they are coarser and finer at the same time.

A finer/coarser topology is not a necessarily “better” one. One may allow too many/too few sets to be considered open, thus erasing all the interesting structure of the topology. For instance, the euclidean topology is coarser than the discrete topology and finer than the trivial one.

Hausdorff space

A topological space is Hausdorff if you can split any pair of points into disjoint open sets. That is, for each \(x,y\) there are open sets \(U, V\) containing \(x\) and \(y\) respectively and \(U \cap V = \varnothing\).

Any metric space is obviously Hausdorff. If \(d(x,y) = k\) then we can use open balls \(B_{\frac{k}{2}}(x), B_{\frac{k}{2}}(y)\) to split them.

Most sort of “natural” topologies are Hausdorff but it’s easy to construct counterexamples. Take the topology \(\{ \varnothing, \{1,2,3,4 \}, \{1,2 \}, \{3,4 \} \}\) on the set 1,2,3,4. Clearly we can’t separate \(1\) and \(2\) into disjoint open sets. The trivial topology is not a Hausdorff space either (except if it contains a single element). In general pseudometric spaces are not Hausdorff (unless they are metric too)

Second-countable space

A topological space is second-countable if it has a countable basis.

Strangely enough euclidean space is second-countable. One may think this is not the case because the basis given by the set of all open balls is uncountable. You are sliding the center across \(R^n\) and \(\epsilon\) across all positive numbers after all. However, we can get a countable basis if we restrict these variables to rationals. That is:

It’s trivial that the topology associated with this new basis is coarser than the euclidean one. We need to show that it’s also finer to conclude that this is indeed a basis for euclidean space.

If we can find a “rational” ball contained in \(B_\epsilon(x)\) that includes \(x\) for any arbitrary \(x, \epsilon\) then we are done. The trick is noting that the rationals are dense. We’ll use the infinity norm for this as it’s more comfortable. Take some \(\hat{x} \in \mathbb{Q}^n\) such that \(\| x - \hat{x} \|_\infty = k < \frac{\epsilon}{2}\) ¹ and take \(\hat{\epsilon} \in \mathbb{Q}^+\) such that \(k < \hat{\epsilon} < \epsilon - k\).

Then using the triangle inequality we have:

Thus, \(x \in B_\hat{\epsilon}(\hat{x}) \subseteq B_\epsilon(x)\)

Not every metrizable space is second-countable though. Take the discrete topology on \(R\). As any basis should include all \(\{x\}\) (you can’t do it as the union of bigger stuff), any basis is going to be uncountable.

Disjoint union topology

Let \(X_i\) be a collection disjoint sets with topologies \(\tau_{X_i}\). The disjoint union topology on \(\cup X_i\) is the topology given by

In other words, a set is open if it’s the union of open sets from each of the topologies. A possible objection to this definition: Why aren’t we defining this new topology as \(\cup \tau_i\)? It’s very simple, if we have two open sets then the union should also be open, we won’t get this if we don’t allow mixing different open sets from different topologies. Another way of seeing this is that we are creating a basis by taking the union of each of the bases.

Product topology for finite stuff

It is very useful to construct new spaces by mixing known ones or to “split” a topological space into simpler more easy-to-understand things.

Take a finite collection of spaces \(X_i\), we define the product topology on \(X_1 \times X_2 \times ...\)

It comes much more naturally to think of it in terms of basis, though.

The topology of \(R^2\) is the product topology on \(R \times R\). It’s sort of natural when you think about it. The product of open intervals is an open rectangle and rectangles are a valid basis for \(R^2\).

Going in the other direction, given a product topology \(X_1 \times X_2 \times ...\) we define the projections in the usual manner. The projection on Xi is a function \(\pi_i: X_1 \times X_2 \times ... \rightarrow X_i\) such that

This straightforward definition of the topology of products is easy only for finite products. When we are taking the product of infinitely (not necessarily countable) many things, generalization gets a bit muddy.

Product topology for infinite stuff

The most natural extension would be to do this basis definition but for infinite stuff. That is:

That’s called the box topology. Another, a bit stranger, generalization is

With \(O_i \in \mathcal{B}_i\) for these finite number of indices.

In other words, for this basis you can only choose a finite subset of indexes to be elements of \(B_i\), the rest needs to span the whole \(X_i\). This one is actually used as the extension of the product topology for infinite products.

It’s direct that the product topology and the box topology are the same for finite products, but they differ for infinite ones. The box topology is finer than the product topology.

Take for instance \(R^\omega = R \times R ...\) infinitely many times. Construct an “infinite dimensional” open cube as \((0,1)^\omega = (0,1) \times (0,1) ...\), this set is open under the box topology but not under the product topology.

What’s the point of this more restrictive definition? This extension to infinite products has a desirable property that the one given by the box topology doesn’t. But first, we need to talk about continuity.

Generalizations using topology

Continuous function

Now that we defined topologies we can talk more freely about the concept of continuous functions in a more general sense without limiting ourselves to euclidean space. It’s pretty cool.

The definition extends naturally from the classical case of continuous functions in euclidean spaces.

Let’s first recall what we mean when we say that a function \(f: R^k \rightarrow R^l\) mapping euclidean topologies is continuous:

\(f:R^k \rightarrow R^l\) is continuous if for any \(\epsilon > 0, x\in R^k\) we have that there is a \(\delta > 0\) such that \(f(B_\delta(x)) \subseteq B_\epsilon(f(x))\). In other words, closer than \(\delta\) from \(x\) means closer than \(\epsilon\) from \(f(x)\)

This automatically tells us how to generalize to any pair of metrizable spaces. But we want more, this definition doesn’t work if we can’t define open balls. We want one that works for arbitrary topological spaces.

So, we develop an equivalent formulation

\(f:R^k \rightarrow R^l\) is continuous if for any \(x\in R^k\) and any open set \(U\) containing \(f(x)\) there’s an open set \(V\) containing \(x\) such that \(f(V) \subseteq U\).

This new definition only talks about open sets. We can generalize it to arbitrary topologies. Let \(f:X \rightarrow Y\) be a function from topological spaces with topologies \(\tau_X\) and \(\tau_Y\) respectively.

Note that whether a function is continuous depends not only on the function itself but on the topologies being considered. A function might be continuous for some topologies but not for others.

We can talk about the continuity of arbitrary functions relating arbitrary topological spaces and not just tuples of real numbers. They don’t even need to be metrizable!

Another sometimes practical (but less illuminating) equivalent formulation of continuity is that a function \(f:X \rightarrow Y\) is continuous iff the preimage \(f^{-1}(U)\) of any open set \(U\) of \(Y\) (that is, the set of elements from \(X\) that map inside \(U\), \(f\) is not necessarily invertible) is open.

Composition of continuous functions is continuous

This new definition of continuity carries a lot of the notions. Composition for continuous functions \(f:X \rightarrow Y, g:Y \rightarrow Z\) is continuous.

Tuple of continuous functions is continuous

Take \(f:X \rightarrow Y, g:X \rightarrow Z\) then \(h:X \rightarrow Y \times Z\) defined as \(h(x) = (f(x), g(x))\) is continuous

Infinite tuples of continuous functions. Box topology and Product topology

Let’s extend this to infinite tuples of continuous functions. Take

Looking at the previous proof of continuity, it relied on doing the intersection of open sets. By the definition of a topology, we know that finite intersections are still open, but that doesn’t work for infinite intersections. Take for instance:

So, is the infinite tuple of continuous functions continuous or not? This is where the difference between the box topology and the product topology becomes meaningful. Infinite tuples of continuous functions are continuous under the product topology but they might not be under the box topology

For the product topology, the proof of the previous section still applies. You end up doing the intersection of a finite number of open sets. That is, given one of these basis elements \(U\) ² in the product topology, we get that.

with \(K\) a finite set. So, it’s open. If we tried it using the box topology we get

We can’t necessarily reduce it to a finite number of intersections. So, we don’t know if this is open or not. Indeed, here is an example that shows that it can break.

Take \(f: R \rightarrow R^\omega\), \(f_i: R \rightarrow R\) with \(f_i(t) = it\). That is, a simple multiplication by \(i\). Here are a couple of values of \(f\) :

It’s a linear function. However, it’s not continuous under the box topology!

The values of the \(f(t)\)-sequence grow out of \((-1,1)\) for all values except \(t= 0\) so it also makes intuitive sense that the preimage of the infinite-dimensional open cube only includes \(0\). The “ray” will always end up escaping from the interval for all values different from 0.

Convergence of sequences

Similar to our generalization of continuous functions we can keep going and generalize convergence. First, let’s recall the definition for sequences of real numbers with a small nudge into how we can generalize from it. For sequences of real numbers \(x_i\), we say that it convergence to \(c\) if for any \(B_\epsilon(c)\) there’s a \(n\) such that \(x_i \in B_\epsilon(c)\) for all \(i\geq n\). By using the fact that any open set includes \(c\) if and only if it contains a ball centered at \(c\) we can get an equivalent definition: It converges if, for any open set \(U\) containing \(c\), \(x_i \in U\) for all \(i\geq n\).

Pretty cool right? All these notions of convergences and such disentangled from some distance function. Convergence in arbitrary topologies is not generally unique. For instance, your sequence might be zig-zagging through multiple values and still say that it is converging not only to a point but to many at the same time! This generalization is not as clean. It’s very easy to see that unique convergence happens if and only if the space is Hausdorff. If you always find two points together in open sets then your topology is too coarse to distinguish one point from the other.

Connected

With this new definition of continuity out of the way, we can now play with definitions that use continuous functions. First, let’s define connected spaces. A space is connected if you can’t split it into two disjoint (non-empty) open sets. This may sound similar to the definition Hausdorff space but the key difference is that we are splitting the whole space into these two sets, not an arbitrary pair of points. A space might be Hausdorff independently of it being connected or not.

For instance, the real line is connected and Hausdorff.

Let’s prove that it’s connected. By contradiction, suppose we could find these two open sets \(U, V\). Take \(x\in U,y \in V\) let’s assume \(x < y\) (otherwise we can switch them up). Take \(z\) the infimum element from \([x,y] \cap V\). Obviously \(x\leq z\leq y\)

• If \(z \in V\) then \(x < z \leq y\). Also as \(V\) is open, \(\hat{z} \in V\) for any \(\hat{z}\) closer than some \(\epsilon\) from \(z\). So we can grab one in between \(x\) and \(z\) and we get \(x <\hat{z} < z\) reaching a contradiction because \(\hat{z} \in [x,y] \cap V\) and we assumed \(z\) was the infimum.

• Similarly, if \(z \in U\) then \(x \leq z < y\). Also \(\hat{z} \in U\) for any \(\hat{z}\) closer than some \(\epsilon\) from \(z\). Take one such that \(z < \hat{z} < y\). But \(\hat{z}\) is not an infimun so we must have some \(v \in V\) such that \(z < v < \hat{z} < y\) which is a contradiction because \(v\) is closer than \(\epsilon\) from \(z\), it must be in \(U\) but \(U\) and \(V\) are disjoint.

The discrete topology is Hausdorff but not connected, but that’s a boring example. Here is a more fun one. Take the set:

With the topology coming from the disjoint union of the natural topology (the euclidean) on the real lines. It’s disconnected, as each line is an open set. On the other hand, it’s Hausdorff. Take any two points: if they are on different lines, any interval around each point will do the trick. If they are on the same line we do the usual open interval stuff.

On a more general note, any disjoint union topology is disconnected.

Now, for an example of a connected but not Hausdorff space. Take \(R \times \{a, b \}\) with the pseudo-metric \(d((x,c), (y,d)) = \lvert x - y \rvert\). It’s not Hausdorff because every open set containing \((x,a)\) also has \((x,b)\). So you can’t split those two. It’s sort of two lines that are “infinitesimally close together” (vs infinitely far apart like the previous example). It’s connected, if it weren’t then using the same argument we used to prove that the real line is connected the only candidate split is \(R \times \{a \}\) and \(R \times \{b \}\) but neither of those are open sets.

Lastly, a space that is neither Hausdorff nor connected. It’s easy, grab a non-Hausdorff space and do the disjoint union with some other space.

Path-connected

In order to gather some intuition about this generalization of continuity it’s a good idea to look at properties and examples.

Path-connectedness is a very intuitive concept in everyday life. Two things are path-connected if you can draw a curve from point A to point B, right? The mathematical formulation of this notion is also easy to understand. Given A and B some objects in Euclidean space, to be path-connected means there’s a continuous function \(f: [0,1] \rightarrow R^k\) that starts at A and ends at B. As simple as that. Continuity is required so as to prevent “jumping” inside the curve. If you move \(t\) a small enough amount then \(f(t)\) also moves a small amount.

Following the footsteps of this definition, given a topological space \(U\) we say that \(x,y \in U\) are path-connected if there’s a continuous function \(f: [0,1] \rightarrow U\) (the path) with \(f(0)=x, f(1)=y\).

It’s kind of strange when you think about it, we can talk about topological spaces being path-connected without the intuition of a curve connecting two points. It’s very weird.

Homeomorphisms

Now that we’ve taken care of all these definitions and generalizations we can talk about more fun stuff. Homeomorphisms are a way to characterize topological spaces. A homeomorphic function is a function between topological spaces \(f: X \rightarrow Y\) is a function such that

• \(f\) is bijective
• \(f\) and \(f^{-1}\) are continuous

We say that two spaces are homeomorphic if there’s a homeomorphic function relating them. It’s trivial that homeomorphism is an equivalence relation. It’s symmetric, reflexive and transitive.

Now, what’s the point of this somewhat arbitrary equivalence? It works as the topologist’s notion of two topological spaces representing the same thing. Expressed like this it looks odd but here’s an equivalent formulation that helps explain the intuition behind it:

So homeomorphic spaces are spaces where you can map bijectively both individual points and neighborhoods. Not only are we renaming points, but we are also preserving the topological structure that relates those points. Spelled like this, it’s much more obvious that homeomorphic spaces are going to share any topological properties (those defined using only topological structure). Things like Hausdorff, path-connectedness or convergence are invariant under homeomorphisms.

Homeomorphisms are also useful to go the other way around: check if two spaces do not share some topological properties to conclude that there’s no homeomorphic function relating them.

Examples of homeomorphic and not homeomorphic spaces

Let’s do some examples to get a bit of intuition behind homeomorphic spaces.

Connected and disconnected spaces are not homeomorphic

Take \(X\) disconnected and \(Y\) connected. Let’s assume that they are homeomorphic with function \(f\) linking them.

As \(Y\) is disconnected we have that the disjoint union of open sets \(V_1 \cup V_2 = X\). But then \(f(V_1) \cup f(V_2) = Y\) is a disjoint union of open sets that generates \(Y\) so \(Y\) is disconnected too.

The unit circle is not homeomorphic to an interval

Take the the unit circle \(S^1 = \{ x^2 + y^2 = 1 \}\) and the open interval \(L = (a,b)\). Intuitively it’s easy to see they are not homeomorphic. You make a hole on the circle and it’s still path-connected, you make a hole on the line and you get two disconnected intervals. That’s the main gist of it. Let’s write it a bit more formally:

Let’s assume that they are homeomorphic with function \(f\) linking them. Let’s remove point \((1,0)\) from the unit circle and \(x = f(1,0)\) from the interval. We get \(L - x = (a,x) \cup (x,b)\)

Grab two arbitrary points \(x_1 \in (a,x)\) and \(x_2 \in (x,b)\).

• \[f^{-1}(x_1) = s_1 = ( \cos \theta_1, \sin \theta_1 )\]
• \[f^{-1}(x_2) = s_2 = ( \cos \theta_2, \sin \theta_2 )\]

(with \(\theta_1, \theta_2 \in (0,2\pi)\) ). Here is a path connecting \(s_1\) and \(s_2\) without going through \((1,0)\)

So \(f \circ g\) connects \(x_1\) and \(x_2\) which we know can’t happen.

All unit norms surfaces are homeomorphic

Take two norm \(\| \|_a\) and \(\| \|_b\) with surfaces:

• \[V_a = \{ \| x \|_a = 1 \}\]
• \[V_b = \{ \| x \|_b = 1 \}\]

Think of the unit square or the unit circle. \(V_a\) is homeomorphic to \(V_b\) (under the euclidean topology)

Here’s the homeomorphic function \(f: V_a \rightarrow V_b\) and its inverse. It’s just a normalization.

• \[f(x) = \frac{x}{ \| x \|_b}\]
• \[f^{-1}(x) = \frac{x}{ \| x \|_a}\]

We need to prove that \(f\) is continuous (and \(f^{-1}\) but it’s symmetric, we ignore it). We could argue that \(f\) is continuous because is a composition of continuous operations (away from 0) but here’s a different proof:

Take an open ball \(B^b_\epsilon(f(x))\), we need to find some open neighoborhood \(U\) of \(x\) such that \(f(U) \subseteq B^b_\epsilon(f(x))\). If we find it we are done.

A bit of algebraic tricks, take \(y \in V_a\) such that

Then by the triangle inequality we have

Now, by noting (again from the triangle inequality)

we can bound the last term by \(\frac{\epsilon}{2}\) and end up with

Then we conclude that

Which is what we wanted. Note that here we are using the fact that all norms are equivalent and so \(B^b_{ \frac{\epsilon}{2} \| x \|_b }(x)\) is open in \(V_a\) ( it’s the restriction of set that’s open set in \(R^n - 0\)).

From this, we get that rhombus are homeomorphic to squares, that spheres are homeomorphic to cubes and such.

Scaling, rotations, and translations are homeomorphic

The function \(f(x) = Ax + c\) with invertible \(A\) is clearly a homeomorphic function so all sorts of rotations, translations, and scalings are homeomorphic.

Graphs of continuous functions are homeomorphic to their domain

Let \(g:X \rightarrow Y\) be a continuous function. Take the graph

with the topology induced from the product topology on \(X \times Y\)

Then \(Z\) is homeomorphic to \(X\). Here’s the homeomorphic function relating \(X\) to \(Z\)

• \[f(x) = (x,g(x))\]
• \[\pi_1((x,y)) = x\]

It’s trivial that both are each other inverse (while only looking at \(Z\), not \(X\times Y\) !). And it’s simple to prove that they and both continuous

So all sorts of funny continuous plots on \([0,1]\) are homeomorphic. All continuous paths from \([0,1]\) in any dimensional euclidean space are homeomorphic if you slap the \(t\) indicating the progress into the coordinates and so on.

References

• Topology without tears by Sidney A. Morris. This is the primary source. Awesome book for self-study. Lots of hand-holding. Check the book here

• Topology by James Munkres. Mostly the first few chapters mostly

• Introduction to Smooth Manifolds by John M. Lee. Didn’t use it that much because I didn’t expand into manifolds as much as I wanted, but it’s also a good one and it has a review of topology at the end.

Footnotes

• There’s also one or two proofs taken from the internet but I can’t remember which and where.