# Thread: If closure(A)=closure(B) prove closure(f(A))=closure(f(B))

1. ## If closure(A)=closure(B) prove closure(f(A))=closure(f(B))

Let $\displaystyle f: X\rightarrow Y$ be a continuous map of topological spaces.

1) Let A and B be subsets of X such that the closure of A = closure of B. Prove that the closure of f(A) = closure of f(B).

2) Prove that if A is dense in X and f(X)is dense in Y then f(A) is dense in Y.

These both look fairly self-intuitive but how can we actaully prove them?

2. If $\displaystyle \overline A$ is closure of A, you should have proved that for continuous mapping, $\displaystyle f\left( {\overline A } \right) \subset \left( {\overline {f(A)} } \right)$.
From the given you know that $\displaystyle f\left( {\overline A } \right) = f\left( {\overline B } \right)$.
Now finish it off.

3. Thanks, this is exactly how far I got, but couldn't see anything further. Am I missing something obvious?

4. Originally Posted by Amanda1990 this is exactly how far I got, but couldn't see anything further.
That should be a lesson to all!
If you had posted what you had done then we would not have wasted time and space repeating it. Originally Posted by Amanda1990 Am I missing something obvious?
The closure of a set is the ‘smallest’ closed superset of the set.

5. OK thanks for your help. Fair point but sometimes it's difficult knowing if you're on the right track at all and so writing down "how far you've got" may just be an assortment of completely irrelevant ideas. I don't see however how this proves part 2 of the question, and as far as I can see I'm not missing anything obvious either...

6. Originally Posted by Amanda1990 I don't see however how this proves part 2 of the question, and as far as I can see I'm not missing anything obvious either...
Well to say that a set is dense in another means $\displaystyle \overline A = X$.
So what does that say about $\displaystyle \overline {f(A)}$?

7. Very nice! Thank you.

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