# 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 $\struct {Y, \tau_Y}$.

This needs considerable tedious hard slog to complete it.In particular: Prove the above statementTo 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 `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

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.

$\blacksquare$