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Suppose $f:X \to Y$ is a continuous proper map between locally compact Hausdorff spaces. Are the following results true?

$1$. The map $f$ takes discrete sets to discrete sets.

$2$. If $f$ is injective, then $f$ must be a homeomorphism onto its image.

Edit1:Let $A$ be discrete in $X$ and let $K$ be compact in $Y$ then $f(A) \cap K=f(A \cap f^{-1}(K))$,is finite since $A \cap f^{-1}(K)$ is finite.Hence $f(A)$ is discrete. As there is a counterexample in answer below so can someone please point out the error in my proof?

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    $\begingroup$ for (2): a closed continuous bijection is always a homeomorphism, the closedness is exactly what you need to prove that the inverse is continuous (use the characterization $f$ is continuous iff the preimage of all closed sets is closed). $\endgroup$ – Holymonk May 24 '15 at 15:09
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    $\begingroup$ You cannot say that $A\cap f^{-1}(K)$ is finite. A discrete subset of a compact space need not be finite, but it would be if it were closed and discrete, i.e. a set without accumulation points. $\endgroup$ – Stefan Hamcke May 30 '15 at 15:34
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    $\begingroup$ The example related to @StefanHamcke comment can be found here: math.stackexchange.com/questions/80802/… $\endgroup$ – Dontknowanything May 31 '15 at 8:53
  • $\begingroup$ i may be being thick but why does $f(A) \cap K$ being finite imply $f(A)$ is discrete? $\endgroup$ – AMFS Oct 16 '17 at 16:15
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Observe that a map $f:X\to Y$ to a compactly generated space $Y$ is a closed map if for every compact set $K$ of $Y$, the restriction $f_K:f^{-1}(K)\to K$ is closed. For if $C$ is closed in $X$, then $f(C)\cap K=f(C\cap f^{-1}(K))=f_K(C)$ is closed in $K$, hence $f(C)$ is closed in $Y$.

As a consequence, a proper map $f:X\to Y$ to a compactly generated Hausdorff space is closed. That's because if $K$ is compact in $Y$, then $f_K$ is a map from a compact space to a Hausdorff space, thus closed.

It follows that a proper map $f:X\to Y$ to a locally compact Hausdorff space $Y$ is closed. If $f$ is injective, then it's an embedding. Also note that a subset $B$ of $Y$ intersecting every compact set in a finite set has no accumulation points, i.e. $B$ is closed and discrete. So if $A$ is closed and discrete, then so is $f(A)$, and this can be proven with the argument in your post.

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  • $\begingroup$ Sorry..where did u use that $Y$ is compactly generated? $\endgroup$ – Dontknowanything May 31 '15 at 9:37
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    $\begingroup$ @Learner: Compactly generated means that a set $A\subset Y$ is closed if $A\cap K$ is closed in $K$ for every compact subspace $K$. $\endgroup$ – Stefan Hamcke May 31 '15 at 11:36
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    $\begingroup$ @Learner: And note that every locally compact space, and more generally, every space in which every point has a compact neighborhood, is compactly generated. $\endgroup$ – Stefan Hamcke May 31 '15 at 14:55
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In (1) the answer is negative even for a compact case. Let $X=[0,1]^2$, $Y=[0,1]$, and $f:X\to Y$ be the projection onto first coordinate. Put $$D=\{(1/n,1/n):n\in\Bbb N\}\cup\{(0,1)\}.$$ Then the set $D$ is discrete, but its image $f(D)$ is a convergent sequence $$\{1/n:n\in\Bbb N\}\cup\{0\}.$$

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  • $\begingroup$ I've added a proof.Can you please point out the flaw in my proof? $\endgroup$ – Dontknowanything May 30 '15 at 13:03

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