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I am reading the definition of "metrizability" which states that if there exists a metric $d$ on set $X$ that induces the topology of $X$, then it is metrizable. My question is how can we possibly know what topology is being induced by metric $d$ which is supposed to be a distance function. How can the distance function and collection of open sets be related?

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The collection of open sets consists of all sets which are unions of open $d$-balls. –  Nate Eldredge Oct 12 '13 at 4:53

4 Answers 4

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As Nate pointed out, the definition of the relationship is that you take the topology generated by the open sets, i.e. finite closures and arbitrary unions over all sets of the form $\{x \in \mathbb{R}:\; |x| < \rho\}$. But I don't think that's what you're asking about, since you seem to understand that a metric is induced on a topology, just not why.

Topological intuition takes some time to get accustomed to. You might picture it as the structure which plays an "in-between" role. Think of the set $\mathbb{R}$ as just a set of elements. It has little structure. Now think of $\mathbb{R}$ as a metric space. It now has a geometry, paths of least distance between two points (geodesics -- in this case, straight lines), and a whole bunch of structure associated to the notion that $d(x,y) = |x-y|$. But in between lies structure that takes no notice of distance...

This structure is the topology of your set, $\mathbb{R}$. You know what open sets are, presumably. [If not, think intervals $(a,b)$, or countable disjoint unions of them -- those are all the open sets in $\mathbb{R}$, a non-trivial fact.] The topology doesn't understand the difference between $(-1,1)$ and $(-\infty,\infty)$, at least not beyond different labelling. The reason is that there is a homeomorphism from $(-1,1)$ to $(-\infty,\infty)$ (hint: there is also one from $(-\pi/2,\pi/2)$ to $(-\infty,\infty)$ -- pick a nice trig function).

A homeomorphism is a transformation which preserves the topology of the space. When you think of the topology on a space, think of "up to homeomorphism". It is continuous and has a continuous inverse, and is bijective. Exercise: show that a homeomorphism $f$ takes open sets to open sets.

So what does all this have to do with the metric? First, open sets give you a notion of "closeness." For instance, you can define convergence of a sequence of points: say $x_n$ converges to $x$ if for any open set $A$ containing $x$ there is some $N$ such that $x_n \in A$ for all $n \geq N$.

The idea of things getting smaller and smaller (or open sets $d(x,y) < \delta$) reflects "fine-grain" structure. Topology, first and foremost, gives you a basic "map" of the space. The finer your topology, the better you know where you are. It's like telling you the continent I live in versus telling you the city. On the right you tell me the city.

It's important to note that specifying which open set you're in tells you about where you are, and where you can't be. For instance, if $A = (-\infty,0)$, then $x \notin A$ means that $x$ has to be on the "right" side of $0$, i.e. $x \geq 0$. That's a geometric property. If I then tell you $x \notin (1,\infty)$, you know more or less where $x$ is -- more or less how far you are from, say, $3/4$.

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A more significant consequence of the above property is that it tells you how open sets are "glued together". The definition of a topology tells you which open sets are which and which give you, say, a non-empty intersection (and which don't!). So for instance, if I took the real line $\mathbb{R}$ and considered $\pm\infty$ to be an "open set" (two elements in one open set! that's possible! but you lose a property), then I would effectively be glueing the two ends of of the real line (think of a rope) together, forming a circle (actually I might be getting something technical wrong here, kind of brain-dead, don't quote me on this).

The relationship to the metric is therefore significant, because you know about being "close" to other points, but you're unable to really quantify how close that really is. That's geometry.

An excellent exercise is to prove that $f$ is continuous (in the $\eps-\delta$ sense) if and only if $f^{-1}(U)$ is open for any open set $U$. Once again, continuity is a property of closeness and it should be related to the topology (just like sequences converging). It's truly remarkable how good the definition of a topology and an open set are.

edit: how can I scale images?

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Wonderful explanation! Almost everything is getting clear that I was having trouble with. I really appreciate your help! Thank you –  Rutherford Mark Oct 12 '13 at 5:45

Suppose that $\langle X,d\rangle$ is a metric space. Let

$$\mathscr{B}=\{B(x,r):x\in X\text{ and }r>0\}\;,$$

where $B(x,r)=\{y\in X:d(x,y)<r\}$ is the open ball of radius $r$ centred at $x$. It is not hard to verify that $\mathscr{B}$ is a base for a topology $\tau$ on $X$. By definition the topology $\tau$ generated by the base $\mathscr{B}$ is

$$\tau=\left\{\bigcup\mathscr{U}:\mathscr{U}\subseteq\mathscr{B}\right\}\;,$$

the family of sets that are unions of members of $\mathscr{B}$. In this case that means that $\tau$ contains the sets that are arbitrary unions of open $d$-balls. These are the open sets of the topological space $\langle X,\tau\rangle$, and $\tau$ is the topology on $X$ inducted by the metric $d$.

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But if you fix $x\in X$, don't you just get one open ball centered around it whose radius is less then $r$? And the region covered by the radius is the set of all points $y$ –  Rutherford Mark Oct 12 '13 at 5:07
    
@Rutherford: I don’t understand the question. If $x$ is a point of $X$, for each real $r>0$ there is an open ball $B(x,r)$ of radius $r$ centred at $x$. –  Brian M. Scott Oct 12 '13 at 5:08
    
I am understanding it this way: $\mathscr{B}$ only contains one element. This is because if you fix $x\in X$ then you can only get one open ball around $x$. Here is the picture I have in mind: Imagine a ball of radius $m$. Draw the radius. At $\frac{m}{2}$ you have your point $x$. Now "somehow" draw another circle with radius less then $r$ centerd around $x$. This new circle of radius $r$ is going to have some region around $x$ which are basically all the points $y$. Thus, in this way you will only get one ball centered around $x$. –  Rutherford Mark Oct 12 '13 at 5:17
    
@Rutherford: But $\mathscr{B}$ explicitly contains lots of elements: one for every possible combination of a point $x\in X$ and a positive real number $r$. The notation $\{u:\varphi(u)\}$ is by definition the set of all objects $u$ such that the statement $\varphi(u)$ is true. Here it’s the set of all open balls $B(x,r)$ such that $x$ is a point of $X$ and $r>0$. –  Brian M. Scott Oct 12 '13 at 5:19
    
Here by $r$ I am seeing it as $\epsilon$. which is as small as possible. So, naively speaking, if you have a open ball of radius, say 2 and you can get an open ball of radius 0.5 in it then you will choose the latter because it has to be less then $\epsilon$ –  Rutherford Mark Oct 12 '13 at 5:22

Let $d$ be a metric. Then the topology induced by $d$ is the topology where the open sets are precisely the sets $U$ with the property that for every $u \in U$ there exists an $\epsilon > 0$ such that $B_\epsilon(u) \subset U$, where $B_\epsilon(u)$ denotes the open ball with radius $\epsilon$ and centre $u$.

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I still don't understand what it means by "induced". If you just give me a metric how can I tell you what topology it induces? –  Rutherford Mark Oct 12 '13 at 4:57
    
It induces the topology $\{U : U \: \textrm{is open}\}$, with open as I defined it in my post. –  Arthur Oct 12 '13 at 5:00

What book are you reading? It must have been carelessly written, if the definition of "metrizable space" is not preceded by a definition of "topology induced by a metric". Here is the missing definition: A set $U$ is open in the topology induced by a metric $d$ if and only if, for each point $a$ in $U$, there is a real number $\varepsilon\gt0$ such that $\{x:d(x,a)\lt\varepsilon\}\subseteq U$. (In other words, just the same way you define open sets in $\mathbb R^n$ using the ordinary Euclidean metric.) The topology induced by $d$ is just the collection of all open sets, as defined above.

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I am reading online notes on general topology –  Rutherford Mark Oct 12 '13 at 5:23
    
Why don't you provide a link so I can read the same notes? Thanks. –  bof Oct 12 '13 at 5:31
    
I have been jumping around to different resources to understand better. I will definitely provide a link if I manage to find the very first source in which I came across the definition. –  Rutherford Mark Oct 12 '13 at 5:42

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