# Closure of Irreducible Subspace is Irreducible/Proof 2

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## 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

This article, or a section of it, needs explaining.In particular: I believe it's important to state carefully in what space a set is closed in, that is, is it closed in $\struct {S, \tau}$ or in $\struct {Y, \tau_Y}$ or $\struct {Y^-, \tau_{Y^-} }$? While this may be obvious to someone who fully understands exactly what is being done here, it is not necessarily obvious for someone who does not fully understand and is trying to gain this understanding.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining 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 `{{Explain}}` from the code. |

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.

$\blacksquare$