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Let $X$ be a topological space and $U$ be open subset of irreducible component $Z$. Then show that if $U$ doesn't meet any other irreducible component, then $U$ is open in whole space $X$

here is the same question but the answer is not for me.

[Question and my try]

I solve this under condition " $X$ has a finitely many irreducible component".

For example, If $x \notin U$ be any element of $X$, then $x$ is in $Z_x$ , where $Z_x$ is an irreducible component which is not $Z$.

If we denote $U = O \;\cap \;Z$ for some open set $O$ in $X$ , then $U = O \;-\; \cup_{x}Z_x $.

Since the union is finite union , $\cup_{x}Z_x $ is closed set so $U$ is open in $X$.

However, If we don't have condition " $X$ has a finitely many irreducible component" , how can we show this now?

Thank you for your attention.

[Conclusion]

Now, I realize that this is a false statement. That is, The answer of here is also not true. Why are you closing this question without even reading this carefully? Some people voted to 'closed', deleted their comments and ran away..

[Answer to my question]

There is a counter-example. Consider the real number $\mathbb{R}$ which has a usual topology. Then every irreducible component is a singleton set. This is because $\mathbb{R}$ is Hausdorff. Then , any singleton set $\{ a \}$ is an open set of itself $\{ a \}$ and disjoint with all another irreducible components. However, $\{ a \}$ is not an open set of whole space $\mathbb{R}$

hew
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  • @Keen-ameteur That link is exactly what I said above in "here" hyperlink. – hew Oct 13 '20 at 13:07
  • What is an irreducible component of $X$ ? – Henno Brandsma Oct 13 '20 at 13:47
  • @HennoBrandsma , A topological space $X$ is ''reducible'' if it can be written as a union $X = X_1 \cup X_2$ of two closed proper subsets $X_1$ , $X_2$ of $X$. A topological space is ''irreducible'' if it is not reducible. Also, "The irreducible component" means a maximal set along these irreducible sets. – hew Oct 14 '20 at 01:00
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    @Leucippus , Again, That link is exactly what I said above in "here" hyperlink. There is no answer about the proof of direction which I want. That link just said "by basic topology, it is true". Also, since that is an old question link , I make this new question – hew Oct 14 '20 at 01:06

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