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For a topological space $(X,\tau)$, the topology $\tau$ on the set $X$ is a family of subsets called open sets, if $X$, $\emptyset$, any union of the subsets, and any finite intersection of the subsets are in $\tau$. This could be a definition of the term "open sets" with respect to topology.

On the real line, $X = \mathbb{R}$, we say $(0,1)$ is "open" based on the definition above, because $\tau_1 = \{\emptyset, \mathbb{R}, (0,1)\}$ can be a topology. However, we say $[0,1]$ is "not open", though $\tau_2 = \{\emptyset, \mathbb{R}, [0,1]\}$ can also be a topology that satisfies all of the properties. I think $[0,1]$ is not called open just because the $\tau_2$ topology is not so-called a standard topology of real line, correct?

I remember the definition of open set for a metric space $(X,d)$ is, we say a subset $S \subset X$ is open if $\forall x \in S$, $\exists \epsilon > 0$ s.t. $B_{\epsilon}(x):=\{y \in X \;|\; d(x,y) < \epsilon\} \subset S$. Is this exactly the standard topology of real line I am asking for, if we collect all of these subsets of $\mathbb{R}$? In other words, we can define open sets in such a way, and collect all these subsets of $\mathbb{R}$ to make a topology, called standard topology. Am I right?

One more question, it seems natural if we take the power set of $\mathbb{R}$, i.e., the discrete topology instead. Why is it not "standard"?

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  • $\begingroup$ Open is a relative term entirely dependent upon your chosen topology. The discrete topology isn't used because it doesn't give us any useful information and every function would be continuous. $\endgroup$ – Gengar Sep 15 '18 at 0:02
  • $\begingroup$ Also, the standard topology on the real line is the one you described with the metric $d(x,y) = |x-y|$. $\endgroup$ – Gengar Sep 15 '18 at 0:05
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    $\begingroup$ It's also possible to define the standard topology on $\mathbb{R}$ to be the order topology induced by the usual order. (Which of course ends up giving the same topology as the Euclidean metric topology.) $\endgroup$ – Daniel Schepler Sep 15 '18 at 0:23
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    $\begingroup$ Hopefully now that your impressions about what the standard topology is have been confirmed, you can see that $\tau_1$ is not the standard topology on $\mathbb R$ either, even though it happens to agree with the standard topology that $(0,1)$ is open. $\endgroup$ – spaceisdarkgreen Sep 15 '18 at 0:30
  • $\begingroup$ Thank you for the comments. Formally, can we say the standard topology of real line includes all $B_\epsilon(x)$ for all $\epsilon > 0$ and $x \in \mathbb{R}$? $\endgroup$ – Analysis Newbie Sep 15 '18 at 12:33
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The word "standard" should refer to the one found for normed spaces. Topology can be very abstract at first.

Looking at your two examples. In $\tau_1$, $(0,1)$ is open: elements in your topology are precisely those we declare open. Your confusion might come from the following situation: If you open a book, where $\mathbb{R}$ is considered, then one would typically say $(0,1)$ is open and not closed. However, one commonly look at $\mathbb{R}$ in the standard (i.e. metric) topology unless one mentions otherwise.

For $\tau_2$: In here $[0,1]$ is again open. This is not the standard topology by any means of course.

For the final bit: The discrete topology is somewhat one of the trivial ones. This is due to every possible singleton being open, so the topology will be the largest one possible. As such, being "open" in the discrete topology is not a special feature. Therefore, it is very unnatural to consider unless you work with intuitively discrete spaces such as $\mathbb{N}$.

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