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If $f$ is one to one and continuous from $X$ onto $Y$, $A$ and $B$ disjoint closed sets in $X$.

Is the intersection of the closures of $f(A)$ and $f(B)$ empty in $Y$?

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What are your thoughts on your question? – Singhal Mar 4 '14 at 13:27

The answer is no. For example, $f:[0,1)\rightarrow S^1, t\mapsto e^{2\pi i t}$ is continuous and bijective, however the inverse is not continuous (and this is why this works).

Set $A=\{0\}$ and $B=[1/2,1)$. They are closed in $[0,1)$ and disjoint, but $f(A)=\{1\}$ and also $1\in \overline{f(B)}$, because $e^{2\pi i t}\rightarrow 1$ as $t\rightarrow 1-$.

Therefore $\overline{f(A)}\cap\overline{f(B)}=\{1\}\not=\emptyset$.

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Not necessarily. For a counterexample, you'll need a continuous bijection $f \colon X \to Y$ that is not a homeomorphism. As an aside, if $X$ is compact and $Y$ Hausdorff, then such an $f$ is automatically a homeomorphism, so we're also looking for a non-compact $X$ (or a non-Hausdorff $Y$, but that is more difficult to visualize).

Take for instance $X = [0,1)$, $Y = S^1$, $f \colon X \to Y$ defined by $f(t) = (\cos(2 \pi t), \sin(2 \pi t))$. Take $A = [0,1/4]$ and $B = [3/4,1)$. Both $A$ and $B$ are closed in $X$; the image of $B$ is not closed in $Y$; and the closure of $f(A)$ and $f(B)$ both contain $f(0) = (1,0)$.

f from [0,1) to S1

And, just for fun, also an example where $X$ is compact and $Y$ is non-Hausdorff. Take $X = [0,1]$ and $Y$ the unit circle $S^1$ with the point $(1,0)$ doubled. Take $f \colon X \to Y$ defined by $$f(t) = \begin{cases} (1,0)_1 & \text{if $t = 0$} \\ (\cos(2\pi t),\sin(2\pi t)) & \text{if $0 < t < 1$} \\ (1,0)_2 & \text{if $t = 1$} \end{cases}$$ and $A = [0,1/4]$, $B = [3/4,1]$. Then both $A$ and $B$ are closed in $X$, but neither $f(A)$ nor $f(B)$ is closed in $Y$, as the closure of something that contains $(1,0)_1$ also contains $(1,0)_2$ and conversely. So, the closure of both $f(A)$ and $f(B)$ contains $(1,0)_1$ and $(1,0)_2$.

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