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

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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 $\varepsilon&gt;0$ such that $B(x,\varepsilon)\subseteq U$.


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