# Connected set on topology - Contradiction

I have a major problem in understanding connected set!

Consider the set $X = (0,1]$. Now lets describe the following sets which can be considered as the basis for the topology on set $X$.

1) For all $b\in X$, all sets of the form $(0, b]$

2) For all $b\in X$, all sets of the form $(b, 1]$

Clearly above 2 groups of sets together can form the basis of the topology on set $X$. Now clearly we can $(0, x]\cup(x,1]$ is the disjoint union of the set $X$.

In the textbook it is written $(0,1]$ as connected BUT we can see it can be expressed as disjoint unions of open sets as above, Hence not connected! Kindly help.

PS: You can argue that sets of $(0, x]$ nature are not open, but they form the topology on $X$, hence they are open.

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Why so complicated? Just say every set is open - that's a topology too. Now only 1 point set is connected... –  user2345215 Mar 18 at 23:37
That did not answer my question! I just showed that the set is not connected by expressing the set as the union of 2 disjoint open sets. –  user135317 Mar 18 at 23:38
You know... connectedness is not about sets, it's about topological spaces. –  user2345215 Mar 18 at 23:39
And I am talking about the topological space on $X$. I have just defined a topology on $X$ and then advanced my argument. –  user135317 Mar 18 at 23:40
yes but i think what user2345215 means is that being connected or not depends on what topology you use. –  Jack Mar 18 at 23:51

Note that the topology that you have defined is not the standard topology on the interval $(0,1]$.

In the standard topology, as a subset of $\Bbb R$, or more concretely, the topology generated by the intervals $(p,q)$ where $p,q\in\Bbb Q$, as well $(p,1]$ where $p\in\Bbb Q$; in that topology the unit interval is connected.

But you changed the topology, so it might be that the space is no longer connected, or not Hausdorff anymore, and so on and so forth.

By the way, it should be pointed that your defined basis is in fact a pre-basis. The intersection of $(0,\frac12]$ and $(\frac13,1]$ is $(\frac13,\frac12]$ which does not contain any sets of the suggested basis.

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That was very helpful! Thanks! –  user135317 Mar 18 at 23:58
You're most welcome. –  Asaf Karagila Mar 18 at 23:59
Hello Asaf! would you be kind enough to see my next question, and if possible shed some light. math.stackexchange.com/questions/718265/… –  user135317 Mar 19 at 14:01
Sorry, that seems way out of my jurisdiction... Me and algebraic topology rarely mix without pain. :-) –  Asaf Karagila Mar 19 at 14:26

You should be carefull, when you say that a set is connected, you are supposed to know what is the topology of the set that you're working. So, if you see the set $(0,1]$as a subset of $\mathbb{R}$ , then it is conected because its open sets are the the set of the form: $X\cap (0,1]$ where $X$ is an open set of $\mathbb{R}$, so your exemple is not correct if the topogy is the topology of $\mathbb{R}$, because the open sets on $\mathbb{R}$ are the open intervals, and the set $[b,1)$ can not be open in this topology.

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Thanks math_man! –  user135317 Mar 19 at 0:02
That's my pleasure! –  math_man Mar 19 at 0:05