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Today for example the teacher ask us to nejdeme next example, but none of us knew, so we left for example homework, but try again but I can not solve the example, so please someone help me, the example is next

Show that if $\{x\}$ is an open set in $X$, for all $x\in X$, then all subsets of $X$ is open in $X$.

Plase help me. Thanky very much. Thanks for your anwers

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    $\begingroup$ Any set is a union of singleton sets. And any union of open sets is open by definition of a topological space. $\endgroup$
    – Dan Shved
    Commented Nov 8, 2013 at 10:43
  • $\begingroup$ That is, if you define topological spaces using open sets. If not, then please clarify what definition of a topological space (and of an open set) you use in your course. $\endgroup$
    – Dan Shved
    Commented Nov 8, 2013 at 10:45
  • $\begingroup$ thank you, but if you can please your question to be detailed because we now learn about these things, my menu to be clearer, please, thank you $\endgroup$
    – Madrit Zhaku
    Commented Nov 8, 2013 at 10:50
  • $\begingroup$ Why down vote, if you tell me please $\endgroup$
    – Madrit Zhaku
    Commented Nov 8, 2013 at 11:08
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    $\begingroup$ MadritZhaku, your question doesn't reflect your attempts to understand the problem. If you could show us, for example, that you know what an open set is, and explain why in this case you can't show that every subset is open, that would go a long way to improving it. $\endgroup$
    – Jonathan Y.
    Commented Nov 8, 2013 at 11:11

2 Answers 2

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HINT: Recall that the definition of an open set in a metric space is as follows.

$U$ is open if for every $x\in U$ there exists some $\varepsilon>0$ such that $B(x,\varepsilon)\subseteq U$.

Now use the assumption that every singleton is open, to conclude that every set is open.

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In a metric space $(X,d)$ every singleton set $\{x\}$ is a closed set. To verify it take any point $y \in X, y \neq x$. Take the distance $d(x,y)$ and see if you are getting an open ball about $y$ to prove $X - \{x\}$ is open.

You can write any subset of $X$ as an union of singleton set. Singleton set is open in our matric space.

We have a result that arbitrary union of open set is open. To prove it, take an arbitrary union of open set and see if you can construct open ball about the point which is contained in the arbitrary union.

So any set in the space will be open.

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  • $\begingroup$ to you before but this is above written @Dan Shved $\endgroup$
    – Madrit Zhaku
    Commented Nov 8, 2013 at 10:59
  • $\begingroup$ Why down vote? Please show reason. $\endgroup$
    – Supriyo
    Commented Nov 8, 2013 at 11:06
  • $\begingroup$ @JonathanY. he has tagged his question "metric space". So metric space is a factor here. He may not know general topology, so I have used metric space only. $\endgroup$
    – Supriyo
    Commented Nov 8, 2013 at 11:11
  • $\begingroup$ Anywhere I did not use Hausdorff property. I like to use only the open ball definition of an open set in a metric space. What is the "crux of the matter"? $\endgroup$
    – Supriyo
    Commented Nov 8, 2013 at 11:18
  • $\begingroup$ There is some extra words. Sorry. $\endgroup$
    – Supriyo
    Commented Nov 8, 2013 at 11:25

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