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  1. If $f$ is a continuous function from $X$ to $Y$ and $A$ is a subset of $X$ then is it true that $x \in A^d$ implies $f(x) \in (f(A))^d$? Here $A^d$ is the derived set of $A$.

  2. If $f$ is a continuous function from $X$ to $Y$ and $X$ is a compact space,then show that $f$ is bounded and attains its bounds.

  3. $A$ and $B$ are two compact subsets of a Hausdorff space $Y$.

    i. Union of $A$ and $B$ is compact in $Y$:

    $A$ and $B$ are closed and hence the union is closed,so it is compact.

    ii. Intersection of $A$ and $B$ is compact in $Y$:

    $A$ and $B$ are closed and hence the intersection is closed,so it is compact.

    iii. $\operatorname{fr}(A)$ is compact:

    $\operatorname{fr}(A)$ is the intersection of $\operatorname{cl}(A)$ and $\operatorname{cl}(Y-A)$. Intersection of two closed sets, hence compact. Am I correct?

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What's the derived set of a set? – Rudy the Reindeer Dec 26 '12 at 7:45
set of limit points of that set. – ccc Dec 26 '12 at 7:52

1.) If $A$ is a subset of $X$ which is mapped to a single closed point in $Y$, then $(f(A))'$ is empty.

Question 2.) only makes sense for a function $f:X\to\mathbb R$. If $X$ is compact, then by continuity of $f$ the image is also compact, hence closed and bounded. So the supremum of $f(X)$ is actually the maximum and there is an $x\in X$ which is mapped to this maximum.

3.) The union of finitely many compact sets is always compact.

In a Hausdorff space compact sets are closed, so the intersection of $A$ and $B$ is a closed subset of the compact set $A$, thus compact.

The boundary is the intersection of cl$(A)$ and cl$(Y-A)$, hence a closed subset of the compact set $A$ and therefore also compact.

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  1. Try using the contrapositive to prove your result ($not$ $f(x) \in [f(A)]^d$ implies $not$ $x \in A^d$), and use the definition of continuity that says that the preimage of every open set is open. Be sure to correctly negate the definition of a point that's $not$ in the derived set.

  2. Using the same definition of continuity as above, take any open cover of $f(X)$, and then the preimages of each of the open sets in that cover, and use the definitions of continuity and compactness to conclude what you want.

  3. i,ii,iii. Have you proved the Heine Borel theorem? It seems that you are using that closed+bounded implies compact.

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  1. It is true because if $f$ is continuous and $\{x_n\}_{n\in\mathbb{N}}\subseteq A$ is a sequence converging to $x$ then the sequence $\{f(x_n)\}_{n\in\mathbb{N}}\subseteq f(A)$ converges to $f(x)$.

  2. I think that it is true only when $Y$ is metrizable. Indeed, $f(X)$ is compact because $X$ is and $f$ continuous. If $Y$ is metrizable then $f(X)$ (since compact) is closed in $Y$ and bounded with respect the metric of $Y$.

  3. Finite union of compact is compact. Finite intersection of compact are compact. Compact subspaces of Hausdorff spaces are closed.

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