continuity of a function

Feb 2011
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0
Hello;

I have a question about continuous functions;

Let A={2,4,6} and B={1,3,5}. A and B are two discrete subsets of the real line so that the topology of them is the induced topology of the real line. Define a function f from A to B such that f(2)=1, f(4)=3 and f(6)=5. Is the function f continuous.


In order to prove that a function is continuous, we need to check that the pre-image of any open set in B is open in A. I confused because there is no open set in B here, so what we can say, is it true that the function f is vacuously continuous.

Thank you in advance
 

FernandoRevilla

MHF Hall of Honor
Nov 2010
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Madrid, Spain
In order to prove that a function is continuous, we need to check that the pre-image of any open set in B is open in A. I confused because there is no open set in B here, so what we can say, is it true that the function f is vacuously continuous.
The induced topologies on $A$ and $B$ are respectively $\mathcal{P}(A)$ and $\mathcal{P}(B)$ (parts of $A$ and $B$), so $f^{-1}(G)$ is open, for all $G$ open. This implies that $f$ is continuous.
 
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Mar 2010
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Yes, I believe that's correct - all subsets of A and B are open since they are the intersection of A (or B) and an open set of real numbers. Therefore every function is continuous.

- Hollywood
 
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Feb 2011
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Thank you so much. that means every subset of A or B is open because the intersection between any open set of real numbers with A or B is either empty or a singelton.

Thank you so much again
 
Mar 2010
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290
Not exactly. Every subset of A or B is open because every subset of A or B is the intersection of an open set of real numbers and A or B. So for example {2,4} is the intersection of A and the open interval (1.5,4.5) (or the open set \(\displaystyle (1.99,2.01) \cup (3.99,4.01)\)).

- Hollywood
 
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