# continuous $f:X\rightarrow X'$, $X,X'$ metric spaces properties

True or False, prove or provide a counterexample.

(a) If $B\subseteq X'$ closed, then $f^{-1}(B)\subseteq X$ is closed.
(b) If $B$ is a bounded subset of $X'$, then $f^{-1}(B)$ is a bounded subset of $X$.
(c) If $A\subseteq X$ and $x_0\in X$ and $f(x_0)\in \operatorname{Acc}(f(A))$ then $x_0\in \operatorname{Acc}(A)$.

My attempts:

(a) True.
Let $B$ be a closed set in $X'$. Suppose $x_0\in f^{-1}(B)$. Take $\epsilon>0$ : $B_{\epsilon}(f(x_0))\subset B^c$, which is open since $B$ is closed. Then $\exists\delta$ : $f^{-1}(B_{\epsilon}(f(x_0))\subset f^{-1}(B^c)$ So $f^{-1}(B^c)$ is open and $f^{-1}(B^c)$ is closed.

(b) False.
Counterexample: $X=X'=\mathbb{R}$. Take $f^{-1}(B)=(-\infty,1)$ and $f(x)=\ldots$ not sure if this is actually false.

(c) True.
For $f(x_0)\in \operatorname{Acc}(f(A))$ to be true, $\forall\epsilon >0: B_{\epsilon}(f(x_0))\cap f(A)\neq\emptyset.$ Does that imply that $B_{\epsilon}(x_0)\cap A\neq\emptyset$?

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What is the definition of acc$(A)$? Also, for b) you can notice that any constant function from $\mathbb{R}$ to $\mathbb{R}$ is a counter-example. Since if $f(x)=c$ for all $x\in \mathbb{R}$, then $f^{-1}\{c\}=\mathbb{R}$ but $\{c\}$ is bounded. – T. Eskin Feb 8 '12 at 7:10
$Acc(A)$ is the set of accumulation points of $A$ – Emir Feb 8 '12 at 7:10
For (c) any constant $f\colon \mathbb R \to \mathbb R$ with constant value $c$, say, also is a counterexample. For if e. g. $A = \{0\}$ and $x_0 = 1$, then $c = f(1) \in f(A)$, but $1 \not\in \mathrm{Acc}(A)$ – martini Feb 8 '12 at 7:22
For (a), it is maybe easier to work with complements. Let $B^c$ be the complement of $B$. It is open iff $B$ is closed. The inverse image of $B^c$ is the complement of the inverse image of $B$. So to say inverse image of a closed set is closed is equivalent to saying that the inverse image of an open set is open. – André Nicolas Feb 8 '12 at 7:23

(a) is indeed true, but your proof as it stands will not work, because you picked $x_0 \in f^{-1}(B)$, so $f(x_0)$ will be in $B$, not $B^c$. You want to show $f^{-1}(B)^c$ is open, I think, so you need to pick $x_0$ in that set. Then indeed $f(x_0) \in B^c$, so as this set is open there is some $\epsilon > 0$ such that $B_{\epsilon}(f(x_0) \subset B^c$, as you then claim. Then you do not use the actual $\delta > 0$, but you probably mean to apply continuity of $f$ at $x_0$, so there is a $\delta > 0$ such that $f( B_{\delta}(x_0) ) \subset B_{\epsilon}(f(x_0))$. Then you skip a minor step: note that we now know that for every $x' \in B_{\delta}(x_0)$, $f(x')$ is in $B_{\epsilon}(f(x_0)) \subset B^c$, so that indeed $x' \notin f^{-1}(B)$, and this shows that indeed, $B_{\delta}(x_0) \subset f^{-1}(B)^c$, showing that the latter set is open, because every point in it is an interior point.
For (b) consider a constant map from say $\mathbb{R}$ to itself. The image is a singleton $B = \{c\}$, and the inverse image of $B$ is all of $\mathbb{R}$, unbounded.
For (c), the opposite is true: $x_0 \in \mbox{Acc}(A)$ implies $f(x_0) \in \mbox{Acc}(f(A))$, and this property characterizes continuity for $f$ (when this holds for all subsets $A$ and all points $x_0$). As it stands it fails: try the function $f(x) = x$ from $\mathbb{R}$ in the discrete metric ($d(x,y) = 1$ for $x \neq y$, $0$ otherwise) to $\mathbb{R}$ in the standard metric. Then $A = (0,1)$ has only $A$ as its accumulation points, as all sets are closed in the discrete metric, but in the image it has $[0,1]$ as its accumulation points, so taking $x_0 = 0$ or $1$ will be a counterexample.