As i am reading up on introduction to point set topology, i saw this example but they did not provide full details. Please help me take a look and see if it is correct! Thanks!

We have to make a good gauge and we pick epsilon $\epsilon$ to be either $x$ or $1-x$, whichever is smaller. This means $\epsilon \leq 0.5$.

Now we need to show that $(x-\epsilon,x+\epsilon)$ is a subset of $G$. Essentially this means that we show for any $u \in (x-\epsilon,x+\epsilon)$, then $u \in G$.

WLOG, we pick $\epsilon = x$, as the other case will be the same.

Now since we know $|u-x| < \epsilon$, we thus have $$|u-x| < \epsilon \Rightarrow -\epsilon < u -x < \epsilon \Rightarrow 0 < u < 2x \leq 1$$

Alternatively, we know from the beginning that $x-\epsilon < u < x+\epsilon $ and we can work from here as well.

This completes the proof as we have shown $u$ is indeed in $G$ for all $u$.

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    $\begingroup$ If I were grading, I'd want a bit more detail/clarification in the last two lines: First, it's polite to the reader to provide the step $|u - x| < \epsilon$ if and only if $-\epsilon < u - x < \epsilon$. Second, I'd suggest clarifying "...$u$ in indeed in $G$ for all $u leq 0.5$." You did this earlier ("WLOG..."), but it's polite to the reader to make sentences correct even out of context. [...] $\endgroup$ – Andrew D. Hwang Aug 5 '17 at 10:53
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    $\begingroup$ Separately, when you write "Now we need to show...", it might be better to express your aim as "First suppose $0 < x \leq 0.5$, and put $\epsilon = x$. We need to show that if $|u - x| < \epsilon$, then $u \in G$." $\endgroup$ – Andrew D. Hwang Aug 5 '17 at 10:55

Your argument is valid. You didn't need the WLOG though; what you wanted to prove was that $u$ is in $(0,1)$. You know that $|u-x|<\epsilon$ therefore, $-\epsilon<u-x<\epsilon$. Adding $x$ yields $$x-\epsilon<u<x+\epsilon\quad (*)$$. Now, $\epsilon=\min\{x,1-x\}$ means that $x+\epsilon\leq x+1-x=1$ and $x-\epsilon\geq x-x=0$. Replacing this to $*$ brings you to $$0<u<1$$ as desired.


Well you can circumvent all of this if you know that the order topology $\mathcal{T}$ on $\mathbb{R}$ has basis $\mathcal{B} = \{(a,b) \ | \ a, b \in \mathbb{R}\}$.

And it it a well-known (and easy to prove) theorem that the order topology on $\mathbb{R}$ and the metric topology (which is the topology you are using) on $\mathbb{R}$ are equivalent.

Therefore any open interval of the form $(a, b)$ is open in $\mathbb{R}$ endowed with either the order or metric topology. Hence $G = \{x \in \mathbb{R} | 0 < x< 1\}$ is open in $\mathbb{R}$ with either of these topologies.

Sidenote: If you're working with $\mathbb{R}$ or any arbitrary set $X$, please mention the topology endowed on the set $X$, after all a topological space is an ordered pair $(X, \mathcal{K})$ where $\mathcal{K}$ is the topology on $X$. The reason I mention this is that $G$ is not universally open in every topology endowed on $\mathbb{R}$, for example $G$ is neither open nor closed in the trivial topology on $\mathbb{R}$, which is given by $T' = \{\mathbb{R}, \emptyset\}$

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    $\begingroup$ The OP said that he/she is "reading up on introduction to point set topology". Knowing what is the basis for the topology of $\mathbb{R}$ and when two topologies are equivalent would mean that he/she has already covered the introduction part... $\endgroup$ – user128787 Aug 5 '17 at 10:42
  • $\begingroup$ In Topology: A First Course by Munkres, what you've just described is what is covered as the very basics of Topology. Basis and comaprable topologies are one of the very first things he talks about. $\endgroup$ – Perturbative Aug 5 '17 at 17:20

G is the open ball with radius 1/2 centered at 1/2.

  • $\begingroup$ But the definition is not "the topology generated by the balls", I think. $\endgroup$ – Henno Brandsma Aug 5 '17 at 10:56
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    $\begingroup$ Yes, but at a point where the OP is asked to prove that an open interval is indeed open, it can't yet be taken for granted that open balls are. He's supposed to go all the way back to the primitive "contains a ball centered on each of its points" definition. $\endgroup$ – hmakholm left over Monica Aug 5 '17 at 10:56
  • $\begingroup$ @HennoBrandsma. So metric space theory does everything backwards? It gives a by point definition of open and closed which then requires proving open and closed are complements, balls are open and generate the open sets? That, perhaps historical approach, is a weak introduction to topology. $\endgroup$ – William Elliot Aug 5 '17 at 20:53
  • $\begingroup$ The pointwise metric topology definition introduces the important notions of interior points and adherent points, which are very important for intuition. Open sets are secondary in metric topology, they lose information. $\endgroup$ – Henno Brandsma Aug 5 '17 at 20:59

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