Closure of Irreducible Subspace is Irreducible
 | Although this article appears correct, it's inelegant. There has to be a better way of doing it. In particular: Can we give the full specification of the topological space? And can we not use $X$, can we use $S$? That is, we talk about $T = \struct {S, \tau}$. Yes I know real grown-up mathematicians treat this suggestion with scorn, but it makes it much more easy to understand what's going on when you are not as familiar with it as a person who thinks topologically every day. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Improve}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $Y \subseteq S$ be a subset of $S$ which is irreducible in $T$.
Then its closure $Y^-$ in $T$ is also irreducible in $T$.
Proof 1
By definition, $Y$ is an irreducible subset of $S$ in $T$ if and only if the subspace $\struct {Y, \tau_Y}$ is an irreducible topological space.
That is, such that two arbitrary non-empty open sets of $\struct {Y, \tau_Y}$ are not disjoint.
The open sets of $T$ in $Y^-$ are the same as the open sets of $\struct {Y, \tau_Y}$.
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Hence by definition of irreducible space, any two open sets in $Y^-$ are not disjoint in $Y^-$.
That is $Y^-$ is also irreducible.
More generally, we can also show that if $Y^-$ is irreducible for a subset $Y \subseteq S$, then $Y$ is also irreducible in $T$.
Aiming for a contradiction, suppose $Y$ is not irreducible.
Then there exist two proper subsets $Y_1$, $Y_2$ of $Y$ which are closed in $Y$ such that $Y = Y_1 \cup Y_2$.
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Then:
- $Y^- = {Y_1}^- \cup {Y_2}^-$
which contradicts the assumption.
$\blacksquare$
Proof 2
Observe that for each subset $V \subseteq Y^-$ which is closed in $T$:
- $(1): \quad Y^- \setminus V \ne \O \implies Y \setminus V \ne \O$
Indeed:
| \(\ds Y \setminus V\) | \(=\) | \(\ds \O\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds Y\) | \(\subseteq\) | \(\ds V\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds Y^-\) | \(\subseteq\) | \(\ds V\) | Closure of Subset of Closed Set of Topological Space is Subset | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds Y^- \setminus V\) | \(=\) | \(\ds \O\) |
Aiming for a contradiction, suppose $Y^-$ is not irreducible.
That is, there exist proper subsets $V_1, V_2$ of $Y^-$ which are closed subsets in $T$ such that:
- $Y^- = V_1 \cup V_2$
Then, from $(1)$:
- $Y \setminus V_1 \ne \O$
and:
- $Y \setminus V_2 \ne \O$
In particular:
- $V_1 \cap Y \subsetneqq Y$
and:
- $V_2 \cap Y \subsetneqq Y$
Because:
- $Y = \paren {V_1 \cap Y} \cup \paren {V_2 \cap Y}$
we have also:
- $V_1 \cap Y \ne \O$
and:
- $V_2 \cap Y \ne \O$
Therefore $V_1 \cap Y$ and $V_2 \cap Y$ are proper closed subsets of $Y$, that form a cover of $Y$.
This contradicts the fact that $Y$ is irreducible.
$\blacksquare$