If S is compact and S= $A\times B$ product of two spaces. Is that enough to state that continuous image of a compact space is compact to state A is compact if we define $f: S\rightarrow$ A is continuous ?
Asked
Active
Viewed 52 times
-2
-
I don't understand what you are asking. – kimchi lover Jul 16 '17 at 23:14
-
1That statament is true only in metric spaces. In metric space compactness and sequential compactness are equivalent things. – ictibones Jul 16 '17 at 23:15
-
I mean if A×B is compact in M×N. How could I prove that A and B are compact too. – seyfullah Jul 16 '17 at 23:16
-
2The projection $A\times B\to A$ is continuous, and the continuous image of a compact set is compact. – kimchi lover Jul 16 '17 at 23:19
-
for the contiunity can we say (a,b) → (a) then f(a,b) →f(a) ? or ? – seyfullah Jul 16 '17 at 23:24
1 Answers
0
If every sequence $(a_n, b_n)$ in $A \times B$ has a convergent subsequence, then, by definition any sequence in $A$ ( or $B$) has a convergent subsequence. If not, we could "lift" a sequence without a convergent subsequence into the product $ A \times B$, to get a non-convergent subsequence. EDIT: This follows from what I think is the standard definition of convergence of a sequence in a product space: we say $(a_n,b_n ) \rightarrow (a,b)$ iff $a_n \rightarrow a $ and $b_n \rightarrow b$
gary
- 4,027
-
-
Thanks for explaining the downvote. My crystall ball is in the shop and my psychic is vacationing in the islands, so I will not be able to guess what is on your mind anytime soon. – gary Jul 22 '17 at 21:53