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Closed set relative

Julio

Member
Feb 14, 2014
71
Show that if $A$ is closed in $X$ and $B$ is closed in $Y$, then $A\times B$ is closed in $X\times Y.$

One question, to be closed in an set this isn't the same that say 'closed set relative'? Because in the last case we have that:

$A=X\cap F_1$ where $F_1$ is closed. Furthermore, $B=Y\cap F_2$ where $F_2$ is closed. It follows that $X$ and $Y$ are closed. Thus,

$
\begin{eqnarray*}
A\times B &=& (X\cap F_1)\times (Y\cap F_2)\\
&=& (X\times Y)\cap (F_1\times F_2)\\
&=& \text{closed in}\, X\times Y
\end{eqnarray*}
$

Finally, at $X$ and $Y$ are closed, then $X\times Y$ is closed. Analogous, $F_1\times F_2$ is closed, because $F_1,F_2$ are closed.
 

Julio

Member
Feb 14, 2014
71
Hello :), can help me?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
You didn't ask a question or pose a problem. What help do you want?

"[FONT=MathJax_Math]A[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]X[/FONT][FONT=MathJax_Main]∩[/FONT][FONT=MathJax_Math]F[/FONT][FONT=MathJax_Main]1[/FONT]A=X∩F1 where [FONT=MathJax_Math]F[/FONT][FONT=MathJax_Main]1[/FONT] is closed. Furthermore, [FONT=MathJax_Math]B[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math]Y[/FONT][FONT=MathJax_Main]∩[/FONT][FONT=MathJax_Math]F[/FONT][FONT=MathJax_Main]2[/FONT]where [FONT=MathJax_Math]F[/FONT][FONT=MathJax_Main]2[/FONT] is closed. It follows that [FONT=MathJax_Math]X[/FONT] and [FONT=MathJax_Math]Y[/FONT] are closed."

No, it doesn't. A set, A, can be "closed in" another set, X, whether X is closed or not.