I am currently studying an introductory book to complex analysis. Within the first chaper the complex numbers are presented in various ways aswell as their structure regarding to different points of view $($e.g. as a vector space, in polar form, as group, etc.$)$.

However, I am having a hard time understanding the concept of topology which is also used there. Especially I am not quite sure whether I have fully understood the idea behind the proofs. Therefore $-$ and moreover in order to pratice proof writing $-$ I picked out a theorem and tried to prove it all by myself. Nevertheless I am not sure if I have managed to work out a sufficient proof or if I missed a crucial point.

Currently I am dealing with topological spaces, metric spaces, and their properties aswell as their relations. Firstly define a topological spaces as a metric space:

All metric spaces are topological spaces
Every metric space $X$ is a topological space, where an open set is defined as the following: a set $M\subset X$ is called "open" if there exists for every $x\in X$ an $\varepsilon>0$, such that the "$\varepsilon$-environment" $$U_{\varepsilon}(x_0):=\{x\in X:d(x,x_0)<\varepsilon\}$$ is completely within $M$.

Here $d(x,y)$ denotes the specific metric of the metric space. Secondly there is a so-called "Hausdorff property" stated as:

"Hausdorff property"
A topological space $X$ is a Hausdorff-space, if there exist for every distinct pair of points $x\ne y$ enviroments $U=U(x)$ and $V=V(y)$ such that $U\cap V=\emptyset$.

My aim is to prove that every metric space satisfies the "Hausdorff-property".


Assume there exists an point $y$ such that $y\in U_{\varepsilon}(x)$ and $y\in V_{\varepsilon'}(x')$ where $x\ne x'$. By definition the enviroments $U_{\varepsilon}$ and $V_{\varepsilon'}$ are defined as

$$\begin{align} U_{\varepsilon}(x)&=\{y\in X:d(y,x)<\varepsilon\}\\ V_{\varepsilon'}(x')&=\{y\in X:d(y,x')<\varepsilon'\} \end{align}$$

From $x\ne x'$ we can conclude that $d(x,x')>0$. Furthermore the triangle inequality states that $d(x,x')\le d(x,y)+d(x',y)$. Consequently this leads to

$$\begin{align} d(x,x')&\le d(x,y)+d(x',y)\\ 0<d(x,x')&\le d(x,y)+d(x',y)\\ 0&< d(x,y)+d(x',y)\\ 0&< d(x,y)+d(x',y)<\varepsilon+\varepsilon' \end{align}$$

Since $\varepsilon,\varepsilon'$ are choosen to be small numbers $($?$)$ this further yields to the contradiction


From where can conclude that $d(x,y)=d(x',y)=0$. Since $d(x,y)$ only equals $0$ iff $x=y$ it follows that $x'=x=y$ which contradicts our assumption.$\small\square$

As marked within my proof I am not sure whether my argumentation about the $\varepsilon,\varepsilon'$ is valid. Anyway I do not know how to finish the proof otherwise.

Is this proof legitimate or not? In the case this is right, what can be improved? In the case it is not, where have I gone wrong? I would appreciate if you could follow my own attempt, i.e. by using the metric of the space in order to prove the "Hausdorff-property". Anyway if this is an impractical approach I am open for different ways of showing the given thereom.

Thanks in advance!

  • 1
    $\begingroup$ Your idea is correct. But you will have to "choose" such $\epsilon$ and $\epsilon '$. You actually have to give them. Can you think of their values? Can you see how these $\epsilon$ environments actually look like in metric space? Try to visualise it, you may find such $\epsilon$ and $\epsilon '$. Another hint is that you need not even find two values. One value will be sufficient. $\endgroup$ – Aniruddha Deshmukh Nov 28 '18 at 18:41
  • $\begingroup$ By the way, do you already know that open balls $U_\epsilon(x)$ are indeed open? It's an immediate consequence of the meric axioms, but you need to show that it contains a ball around each of its points, not just $x$ $\endgroup$ – Hagen von Eitzen Nov 28 '18 at 19:09
  • $\begingroup$ @HagenvonEitzen I would rather see this as a definition in this case than an assumption. The inclusion of the first statement was just to add the connection of the metric space and the topological space by the "$\varepsilon$-enviroment" which I tried to use. Would it be an important point in order to just show that a metric space fulfills the "Hausdorff property", or can it be neglected in this case? $\endgroup$ – mrtaurho Nov 28 '18 at 19:16

The main issue with your proof is that you haven't actually said what $\varepsilon$ and $\varepsilon'$ are before they're used, and then later on you say they're "small numbers", but again this is too vague.

You then conclude that $d(x,y) + d(x',y) < 0$, but this conclusion is erroneous for two reasons:

  • $\varepsilon$ and $\varepsilon'$ are (or, rather, should be) fixed from the start, and so you can't jump from $d(x,y) + d(x',y) < \varepsilon + \varepsilon'$ to this conclusion.
  • Even if you had proved that $d(x,y) + d(x',y) < \varepsilon + \varepsilon'$ for all $\varepsilon,\varepsilon' > 0$ (which you can't, since they're supposed to be fixed), the conclusion that you'd actually end up drawing is that $d(x,y) + d(x',y) \le 0$. But this is a moot point, as I mentioned.

Here's how I'd suggest to fix your proof. Take $x,x' \in X$ with $x \ne x'$. Then $d(x,x') > 0$, and so you can let $\varepsilon = \dfrac{d(x,x')}{2}$, which is positive (and is now fixed!).

Use a proof strategy similar to what you already tried, in order to prove that $U_{\varepsilon}(x)$ and $U_{\varepsilon}(x')$ are disjoint open neighbourhoods of $x$ and $x'$, respectively.

  • $\begingroup$ Just as a clarification for myself: By actually choosing $\varepsilon=\varepsilon'=\frac12d(x,x')$ I will get the inequality chain $$d(x,x')\le d(x,y)+d(x',y)< d(x,x')$$ but how exactly can I conclude from hereon that this contradicts my assumption? I mean indeed this seems to me already to be a contradiction since the sum of the two inner distances cannot be less or equal and greater than the distance between $x$ and $x'$ at the same time from where I can conclude that they both have to be zero. Or am I mistaken? $\endgroup$ – mrtaurho Nov 28 '18 at 18:51
  • $\begingroup$ That's not the inequality chain you'll get. Assume $y \in U_{\varepsilon}(x) \cap U_{\varepsilon}(x')$. Then $$d(x,x') \le d(x,y) + d(y,x') < \varepsilon + \varepsilon = 2\varepsilon = d(x,x')$$ So the contradiction you get is that $d(x,x') < d(x,x')$. $\endgroup$ – Clive Newstead Nov 28 '18 at 18:54
  • $\begingroup$ It was just a typo, sorry for that. Nevertheless now it is clear. I am only a little bit curious about the choice of $\varepsilon$ and $\varepsilon'$; it seems to me like some kind of "cheating" to choose the value in order to cause this contradiction. How may on justify this choice? $\endgroup$ – mrtaurho Nov 28 '18 at 18:57
  • $\begingroup$ @mrtaurho: What makes you say it's cheating? The definition of Hausdorff says that for every pair $x,x' \in X$ with $x \ne x'$, there are disjoint open neighbourhoods $U$ of $x$ and $V$ of $x'$. To prove this, you let $x,x' \in X$ be arbitrary elements with $x \ne x'$, and then in terms of $x$ and $x'$, all you have to do is find a neighbourhood of $x$ and a neighbourhood of $x'$ that are disjoint. Any two open balls that are small enough that they don't overlap will do, so we pick the radius of the open balls to be (at most) half the distance between $x$ and $x'$, so that they don't overlap. $\endgroup$ – Clive Newstead Nov 28 '18 at 19:00
  • $\begingroup$ If what you're worried about is that $\varepsilon$ is defined in terms of $x$ and $x'$, that's just what proofs of statements of the form $\forall \cdots \exists \cdots$ look like: in terms of the $\forall$-quantified variable(s), find a single thing (or, in this case, a single pair of things) making the statement true. $\endgroup$ – Clive Newstead Nov 28 '18 at 19:02

You wrote that $\varepsilon,\varepsilon'$ were chosen to be small numbers, but you did not tell us which choice was that. Therefore, you proved nothing.

Simply take $\displaystyle\varepsilon=\varepsilon'=\frac{d(x,x')}2$. That will do: $U_\varepsilon(x)\cap U_{\varepsilon'}(x')=\emptyset.$


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