clousre

**flower3** · May 28th 2009, 10:19 AM

Suppose that $\displaystyle \tau 1 \mbox and \quad \tau2 \mbox \quad are \quad \quad \quad topologies \quad on \quad a set X \quad with \quad \tau1 \subset \tau2 . \mbox \quad if A \subset X \quad$

$\mbox prove \quad that : \quad \overline{A}^2 \subset \overline {A}^1 \mbox \quad where \overline {A}^1 \mbox is \quad the \quad clousre$ $\quad of A \quad in (X, \tau1 ) \quad \overline {A}^2 \mbox is \quad the \quad clousre \quad of A \quad in (X,\tau2)$

**Jose27** · May 28th 2009, 12:40 PM

If $A$ is an open subset in $\tau_1$ then $A$ is open in $\tau_2$ . So, if $B$ is closed in $\tau_1$ then $B$ is closed in $\tau_2$ . Now you define the closure of $A$ in $\tau_i$ as the intersection of all closed subsets of $(X, \tau_i)$ such that they contain $A$ , but since closed sets in $\tau_1$ are closed in $\tau_2$ then the closure in $\tau_2$ has more elements to intersect and thus is contained in the closure in $\tau_1$ .

Well, that is an overview of the proof. I leave the details to you.

**Plato** · May 28th 2009, 01:13 PM

Here is a different approach.
$x \in \overline A$ if and only if each for each open set $O$ , $x \in O\, \Rightarrow \,O \cap A \ne \emptyset$ .
But $O \in \tau _1 \; \Rightarrow \;O \in \tau _2$ .

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