# What is a Topological Space?

Following on from my post last week about metric spaces, I thought today I would describe what a topological space is.

### Definition

A Topological Space (X, τ) consists of a set C and a set (the topology) τ of subsets of X such that:

1. ∅, X ∈ U;
2. If Ui ∈ τ for all i ∈ I, then ;
3. If U1, U2 ∈ τ, then U1 ∩ U2 ∈ τ.

The closure property (#3) is true for all finite intersections.

The elements of X are the points and the elements of τ are the open subsets of X.

Note that a subset Y of X in a topological space (X, τ) is called closed if X\Y is open, so we can describe a topology on a set X by specifying the closed sets in X which satisfy:

• ∅, X are closed;
• If Fi is closed for all i ∈ I, then so is ;
• If F1, F2 are closed, then so is F1 ∪ F2.

### Metric Topologies

A topology can be induced by a metric space (X,d); These are called metric topologies.

For example, the discrete metric on a set X gives rise to the discrete topology in which every subset in X is open, i.e. τ is the power set of X.

### Example of Non-Metric Topologies

1. Let X be a set with at least two elements, and τ := {X,∅}. This is called the indiscrete topology.
2. Let X be any uncountable set, such as ℝ or C and τ := {∅}∪{Y ⊂ X: X\Y is countable}. This is called the co-countable topology.
3. Let X be any infinite set and τ := {∅}∪{Y ⊂ X: X\Y finite}. This is called the co-finite topology. If X = R or C then this is known as the Zariski toplogy.

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# VIDEO: Napkin Ring Problem

If you were to core a sphere (remove a cylinder from it), you are left with a shape that looks like a napkin ring. This is a “bizarre” shape, as if you have two napkin rings with the same height, they will have the same volume regardless of the size of the initial spheres that they came from. How do you prove this?

Here’s a few hints to try and solve it yourself before watching the Vsauce video below which reveals the answer:

• There are a few variables that need to be found: the height of the napkin ring, the radius of the starting sphere and the radius of the cylinder. Using these variables you can find a volume equation.
• You don’t need to find the volume of the whole napkin ring in one go. This is because, as the two napkin rings have to be the same height, it’s enough to show that any slice of the napkin rings has to have the same area. If every pair of slices has the same area, then the napkin rings have the same volume.

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# What is a Metric Space?

Given a set X, it is often helpful to define a notion of distance between points in this set. This distance function is called a metric.

### Definition

metric space is a pair (X, d), where X is a set (the space) and is a function d: X × X → R (the metric) such that, for all x,y,z we have:

• Non-negativity: d(x,y) ≥ 0, where d(x,y) = 0 iff x = y;
• Symmetry: d(x,y) = d(y,x);
• Triangle Inequality: d(x,z) ≤ d(x,y) + d(y,z).

### Examples

Euclidean Metric: This is the usual metric used, and one that you are probably all familiar with.

Let X = R^n and

This is the usual notion of distance that we have in the R^n vector space. Using the Cauchy-Schwarz inequality, we can easily show that this is a metric by showing that it satisfies the 3 axioms above.

Discrete Metric:

Let X be a set and

The first two axioms are trivially satisfied, so to show that the above is indeed a metric, we just have to show that the third is also satisfied. We can prove this by exhaustion:

Since the distance function can only output 0 or 1, f(x,z) can be 0 or 1, while d(x,y) + d(y,z) can be 0, 1 or 2. For the third axiom to fail, it must be the case that the RHS < LHS, which can only happen if the RHS is 0. However, if the RHS is 0, we must have the x = y = z, so the LHS is also 0, showing us that the third axiom is always satisfied.

British Railway Metric:

Let X = R^2 and define

To explain the name of this metric, think of Britain with London as the origin. Then in a less than ideal railway system, all trains go through London. For example, if you want to go from Oxford to Cambridge, then you would go from Oxford to London, and then from London to Cambridge. The resulting distance travelled would be the distance of both these journeys added up.

The exception is when the two destinations lie along the same line. In this case, you can take the train from one to the other without going through London – hence the if x = ky clause.

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# Noether’s Theorem

Today I thought I’d write a blog post about an interesting theorem I learnt whilst studying my Variational Principles module – Noether’s Theorem.

To understand Noether’s Theorem, we must first understand what is meant by a symmetry of a functional.

Given

suppose we change the variables by the transformation t –> t*(t) and x –> x*(t) to obtain a new independent variable and a new function. This gives

where α* = t*(α) and β* = t*(β).

If F*[x*] = F[x] for all x, α and β, then this transformation * is called a symmetry.

What is a continuous symmetry?

Intuitively, a continuous symmetry is a symmetry that we can do a bit ofFor example, a rotation is a continuous symmetry, but a reflection is not.

### Noether’s Theorem

Noether’s Theorem – proven by mathematician Emmy Noether in 1915 and published in 1918 – states that every continuous symmetry of F[x] the solutions (i.e. the stationary points of F[x]) will have a corresponding conserved quantity.

### Why?

Consider symmetries that involve only the x variable. Then, up to first order, the symmetry can be written as:

t –> t, x(t) –> x(t) + εh(t)

where h(t) represents the symmetry transformation. As the transformation is a symmetry, we can pick ε to be any small constant number and F[x] does not change, i.e. δF = 0. Also, since x(t) is a stationary point of F[x], we know that if ε is any non-constant, but vanishes at the end-points, then we have δF = 0 again. Combining these two pieces of information, we can show that there is a conserved quantity in the system.

For now, do not make any assumptions about ε. Under the transformation, the change in F[x] is given by

Firstly, consider the case where ε is constant. Then the second integral vanishes and we obtain

So we know that

Now, consider a variable ε that is not constant, but vanishes at the endpoints. Then, as is a solution, we must have that δF = 0. Therefore,

If we integrate the above expression by parts, we get that

Hence the conserved quantity is:

Not all symmetries involve just the x variable, for example we may have a time translation, but we can encode this as a transformation of the x variable only.

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# MATHS BITE: Shoelace Theorem

The Shoelace theorem is a useful formula for finding the area of a polygon when we know the coordinates of its vertices. The formula was described by Meister in 1769, and then by Gauss in 1795.

### Formula

Let’s suppose that a polygon P has vertices (a1, b1), (a2, b2), …, (an, bn), in clockwise order. Then the area of P is given by

$$\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|$$

The name of this theorem comes from the fact that if you were to list the coordinates in a column and mark the pairs to be multiplied, then the image looks like laced-up shoes.

### Proof

(Note: this proof is taken from artofproblemsolving.)

Let $\Omega$ be the set of points that belong to the polygon. Then

$$A=\int_{\Omega}\alpha,$$

where $\alpha=dx\wedge dy$.

Note that the volume form $\alpha$ is an exact form since $d\omega=\alpha$, where

$$\omega=\frac{x\,dy}{2}-\frac{y\,dx}{2}.\label{omega}$$

Substitute this in to give us

$$\int_{\Omega}\alpha=\int_{\Omega}d\omega.$$

and then use Stokes’ theorem (a key theorem in vector calculus) to obtain

$$\int_{\Omega}d\omega=\int_{\partial\Omega}\omega.$$

where

$\partial \Omega=\bigcup A(i)$

and $A(i)$ is the line segment from $(x_i,y_i)$ to $(x_{i+1},y_{i+1})$, i.e.  is the boundary of the polygon.

Next we substitute for $\omega$:

$$\sum_{i=1}^n\int_{A(i)}\omega=\frac{1}{2}\sum_{i=1}^n\int_{A(i)}{x\,dy}-{y\,dx}.$$

Parameterising this expression gives us

$$\frac{1}{2}\sum_{i=1}^n\int_0^1{(x_i+(x_{i+1}-x_i)t)(y_{i+1}-y_i)}-{(y_i+(y_{i+1}-y_i)t)(x_{i+1}-x_i)\,dt}.$$

Then, by integrating this we obtain

$$\frac{1}{2}\sum_{i=1}^n\frac{1}{2}[(x_i+x_{i+1})(y_{i+1}-y_i)- (y_{i}+y_{i+1})(x_{i+1}-x_i)].$$

This then yields, after further manipulation, the shoelace formula:

$$\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i).$$

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# Proof Without Words #3

Proof of the identity

The figure for general n is similar, with n right pyramids, one with an (n-1)-cube of side length xk as its base and height xk for each k=1,….,n.

The derivative of sin is cosine.

From ‘Proof without words‘ by Roger Nelsen

By Sidney H. Kung

By Shirley Wakin

Previous ‘Proof Without Words‘: Part 1 | Part 2

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# Maths Bite: Impossible cube

The impossible cube was invented by M.C. Escher for his 1958 print Belvedere. It is based on the Necker cube, and seems to defy the rules of geometry; on the surface resembles a perspective drawing of a 3D cube, however its features are drawn inconsistently from the way they would be in an actual cube.

The impossible cube draws upon the ambiguity present in a Necker cube illustration, in which a cube is drawn with its edges as line segments, and can be interpreted as being in either of two different three-dimensional orientations. – Wikipedia

Source: kidsmathgamesonline

How would this cube look like in real life? The below video attempts to demonstrate that.

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