"What does it mean for a set to be open in a Topology?" - I have a hard time getting my head around what it means.


Is the set $(0,1]\subset\mathbb{R}$ open in the standard Topology on $\mathbb{R}$ or open in the Topology generated by the basis $B = \{[a,b)\subset \mathbb{R} | a < B \}$ on $\mathbb{R}$?

My answer / idea:

I would think that since the standard topology is generated by open intervals, we could have the interval $(0,2)$, which contains $(0,1]$ - so I guess that it would be "open" in the topology.

For the Topology on the basis B, I would also say that since $(0,1] \subset [0,2)$ it must also be open in that topology.

Though, I have a clear feeling that I have the wrong idea. I know that the interval being simply contained in the Topology, is not an answer.

  • 1
    $\begingroup$ "Open set" is just a synonym for a set that is in the topology. If you want to know if a set $U$ is an open set of a topology, take its basis and try to build $U$ using unions and finite intersections. $\endgroup$ – H. Potter Apr 23 '16 at 18:03
  • $\begingroup$ Your arguments suggest that you believe "Every subset of an open set is open". This is far from being correct (especially as the whole space is always open) $\endgroup$ – Hagen von Eitzen Apr 23 '16 at 18:04
  • $\begingroup$ $(0,1]$ is neither open nor closed in the standard topology. (Same for the other topology, I believe.) $\endgroup$ – Akiva Weinberger Apr 23 '16 at 19:16

Recall the definition. A topological space is a pair $(X,\mathcal{T})$where $X$ is a set and $\mathcal{T}$ (is defined to be the set of open sets,) $\mathcal{T}$ is a collection of subsets of $X$ such that

  1. $\emptyset\in\mathcal{T}$
  2. Any union $A$ of a collection of $A_i\in\mathcal{T}$ is also in $\mathcal{T}$
  3. Any finite intersection $A$ of $A_1,\ldots,A_n\in\mathcal{T}$ is also in $\mathcal{T}$.

Then the topology generated by a set $\mathcal{A}$ is the coarsest ("smallest") topology $\mathcal{T}$ such that $\mathcal{A}\subseteq\mathcal{T}$.

So an open space in the topology genrated by the intervals $B=\{[a,b)\subseteq\Bbb{R}:a<b\}$ has the open sets $A\in\mathcal{T}$ such that $A$ can be obtained via taking arbitrary unions or finite intersections of $B$-intervals. For example, $[0,2)$ is open, as is $\bigcup_{n\ge 1}[0,n)=[0,\infty)$. For a more interesting example, consider the set given by $$\bigcup_{n\ge 2}\,[1/n,1),$$ which is also open, as an arbitrary union of open sets; notice this union is equal to $(0,1)$.

On a side note, $(0,1]$ is in fact not open. To show a set is not open, one way is to show that the complement is not closed. In your example, to show that $(0,1]$ is not open in the topology generated by $B$, I would show that $(-\infty,0]\cup(1,\infty)$ is not closed in the topology generated by $B$.

Lemma. A subset $C$ is closed iff every convergent (in $X$) sequence $x_n\to x$ with points $x_n\in C$ has $x\in C$.

To show that $x_n\to x$ in the topology generated by $B$ it is sufficient to show that for every set in the basis $[a,b)\in B$, that contains $1$, there is some $N$ such that $n\ge N$ implies $x_n\in [a,b)$. Let $x_n=1+1/n$, I claim that $x_n\to 1$ in $(\Bbb{R},B)$. For every base element containing $1$ (i.e. $[1,1+\varepsilon)$), there is an $N$ such that $n\ge N$ implies $1/n<\varepsilon$, i.e. $x_n\in [1,1+\varepsilon)$; but $1\notin(-\infty,0]\cup(1,\infty)$. Hence, $(-\infty,0]\cup(1,\infty)$ is not closed, so $(0,1]$ is not open.

  • $\begingroup$ Thanks for the great response. However, I am not completely sure how to understand it yet. In the last part you show that (0,1] is not open, but is that really relevant? If a set is not open, does that imply that it can't be open in any topology? However, your example does make sense (to me!! :) ) if it is in relation to the standard topology - which is made of open intervals which would make it obvious that a closed set could not be a member of this topology then. However, how would you apply this to my second example with the Topology on R with basis B? $\endgroup$ – Fredrik Apr 26 '16 at 3:47
  • $\begingroup$ Here is how I see it (please do correct me). I would say that setting $a = 1/n$ and $b = 1 + 1/m$, where $n,m$ is greater than some large $N$. Then $(0,1] \subset [a,b)$, but it is not possible to pick an $m$ (unless using the limit) such that $b \neq 1$. From this, I guess that since it is not possible to pick an $b = 1$, then $(0,1]$ is not a member of the topology - thus not not open. $\endgroup$ – Fredrik Apr 26 '16 at 3:58
  • $\begingroup$ @Fredrik I showed that $(0,1]$ is not open in the topology generated by $R$, note that as the neighborhoods of $1$, I used the sets $[1,1+\varepsilon)$ which are members of $R$. To show that $(0,1]$ is not open in the standard topology, you might consider using the neighborhoods $(1-\varepsilon,1+\varepsilon)\ni 1$. Note that since $[1,1+\varepsilon)$ is not open in the standard topology, this argument would not work. And indeed it is NOT true that $(0,1]$ is not open in every topology: in the discrete topology, all subsets of $\Bbb{R}$ are open! $\endgroup$ – Szmagpie Apr 26 '16 at 11:57
  • $\begingroup$ I have corrected a couple of errors and added some detail to that part of my answer. Hope it's clear. $\endgroup$ – Szmagpie Apr 26 '16 at 12:19

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