Can we prove that a bounded closed subset of $\mathbb R^n(n \ge 1)$ is compact without using Axiom of Choice?  This is  which was closed.
120573
Heine-Borel Theorem; If $E \subset \mathbb{R}^k$, then $E$ is compact iff $E$ is closed and bounded.  I have proved 'closed and bounded⇒compact' and 'compact⇒bounded'. (There exists $r\in \mathbb{R}$ such that for every $x\in E$, $|x|&lt;r$)  The proof in Rudin PMA p.40 uses 'countable axiom of choice'  I have googled it and found some proofs, but they all used some weaker form of AC.  Please help me how to show that $compact⇒closed$ in ZF..
99650
$(\to)$ Assume that $S \subseteq R$ is bounded. If $(x_n)$ is a sequence in $S$ , then $(x_n)$ is bounded and thus it has a convergent subsequence.  $(\leftarrow)$ Assume that $S$ is not bounded. Then for any integer $n$ there is an $x_n$ in $S$ such that $|x_n| &gt; n$. Since any convergent sequence is bounded, the sequence $(x_n)$ cannot have a convergent subsequence.  Can someone check to see whether my proof is missing something.  thank you.
96502
Given two sets $S_1$ and $S_2$ in $\mathbb R^n$ define their sum by    $$S_1+S_2=\{x\in\mathbb R^n; x=x_1+x_2, x_1\in S_1, x_2\in S_2\}.$$   Prove that if $S_1$ and $S_2$ are compact, $S_1+S_2$ is also compact.   Prove that the sum of two compact sets in $\mathbb R^n$ is compact.   Compact set is the one which is both bounded and closed. The finite union of closed sets is closed. But union is not the same as defined in the task. I so not know how to proceed. I do understand that I need to show that the resulting set is both bounded and closed, but I do know how to do that.
36804
I am studying functional analysis, and in the setting of normed spaces I have seen the theorem that states that the unit ball is compact iff the space is finite dimensional.  I also saw an exercise:     Let $X$ be finite dimensional, prove that for any non-empty and closed   $C$, $x\in C$ there exist $c\in C$ s.t $||x-c||=d(x,C)$.   The proof started with:     take $c\in C$ and consider  $$ D=\overline{B(x,||x-c||)}\cap C $$      $D$ is closed and bounded hence, since $X$ is finite dimensional, is   compact.   I would be glad if someone could help me with the following two questions:  1) It seems that the theorem I read about the compactness also extends to $X$ is finite dimensional iff $\{x,||x-x_{0}||\leq d\}$ is compact for any $x_{0}\in X,d\in\mathbb{R}^{+}$. is this correct ?  2) $C$ is closed, $\overline{B(x,||x-c||)}$ is also closed, hence so is the intersection. since $\overline{B(x,||x-c||)}$ is bounded then so is the intersection.  And so I get that $D$ is closed and bounded. Why can't I conclude that $D$ is compact without the assumption that $X$ is finite dimensional ?
47799
Suppose $X$ is a non-compact metric space. Show that there exists a continuous function $f: X \rightarrow \mathbb{R}$ such that $f$ does not achieve a maximum.   I proved this assertion as follows:  If $X$ is non-compact, there exists a sequence which has no converging subsequence. Therefore, the set $E$ of elements of this sequence is infinite and has no limit points. Therefore, in the induced topology, $E$ is a countable discrete space. Let $e_n$ be a enumeration of $E$. Define $f: E \rightarrow \mathbb{R}$ as $f(e_n)=n$. Since $E$ is discrete, $f$ is continuous. By Tietze (note that $E$ is closed in $X$), there exists a continuous function $g: X \rightarrow \mathbb{R}$ which extends $f$. But $f$ is unbounded, then so is $g$, and $g$ does not achieve a maximum. $\square$  My first question is: Is everything in the proof fine?  Now, my second question is: Can you provide me a proof which is more elementary? (Not using Tietze, for example. This exercise is just after the definition of compactness. I suppose this proof is not what was intended.)  And my third question is: Does the proposition hold for arbitrary topological spaces? If not, can you provide a counter-example?
15655
Let $f: X \to Y$ be continuous and proper (a map is proper iff the preimage of a compact set is compact). Furthermore, assume that $Y$ is locally compact and Hausdorff (there are various ways of defining local compactness in Hausdorff spaces, but let's say this means each point $y \in Y$ has a local basis of compact neighborhoods).  Prove that $f$ is a closed map.  I know that this proof cannot require much more than a basic topological argument. But there's just something that I'm missing.   We can start with $C \subseteq X$ closed, and then try to show that $Y \setminus F(C)$ is open (for each $q \in Y \setminus F(C)$, we would want to find an open set $V_q$ with $q \in V_q \subseteq Y \setminus F(C)$).   Hints or solutions are greatly appreciated.
122542
Recall that a space $X$ is locally compact if for any point $x$ in $X$, and any neighborhood $U$ of $x$ in $X$,there is a neighborhood $V$ of $x$ in $X$,and a compact subspace $C$ of $X$ such that $x \in V \subset C \subset U.$ Let $X$ be a locally compact space, and let $f : X \rightarrow Y$ be a continuous, closed surjection. Suppose that $f^{-1}(y)$ is compact for each $y \in Y$ . Show that $Y$ is locally compact.
57254
The following proof is based on Rudin's Principles, chapter 2, exercise 26. I wanted to confirm that the axiom of (countable) choice is used in the line in bold, and was curious to know whether this proof can be modified so that it isn't required, and if not, whether there is an entirely different proof which doesn't require choice.  Let $X$ be a metric space in which every infinite subset has a limit point. By previous exercises, we know that $X$ must have a countable base. So every open cover of $X$ has a countable subcover $\{ G_{n} \}$. Assume that $\{ G_{n} \}$ has no finite subcover. Construct a set $E$, containing one element from each set $F_n = (\bigcup_{i=1}^{n}G_i)^c$. Then $E$ is infinite, since every point in it is in some $G_m$, so won't be in $F_{m}$. But then $E$ must have a limit point $y$, which is in contradiction with the fact that $y$ must be in some $G_n$, which is open. Thus $\{ G_{n} \}$ must have a finite subcover.
117928
Let $A \in M_n(\bf C)$ and let   $S=\{ A \in M_n(\bf C) : A=A ^*$ and  $\rho (A) \leq 1 \}$, $\rho(A)$ denotes spectral radius of $A$.     Show that $S$ is compact subset of $M_n(\bf C) \cong \bf C^{n^2} $ with usual topology.    $S$ is subset of $\bf O_n(C)$ which is compact so I think its enough to show that $S$ is closed.Please help.
73741
The question is: let $F_1, ... F_n$ be compact subsets of X. Show that $\cup^{N}_{n=1} F_n$ is compact.   know that a set $ F  \subset X$ is compact if every open cover $\mathcal {G}$ of F contains a finite subcover $\mathcal {H}$. Intuitively, I think that since $F_1, .. F_n$ are compact, they contain finite subcovers and so $\cup^{N}_{n=1} F_n$ at most has finite subcovers. I'm just stuck at trying to formally prove this out.   Help would be much appreciated!
84385
Let $\{K_i\}_{i=1}^{\infty}$ a decreasing sequence of compact and non-empty sets on $\mathbb{R}^n.$ Then $\cap_{i = 1}^{\infty} K_i \neq \emptyset.$  I heard about a proof that take $x_i \in K_i.$ Then I have constructed a sequence. It is bounded because for every $i$, $x_i \in K_1$ that is bounded. Then there is a subsequence that converges. That is where I stopped. What indeed means that a subsequence converges? And why does the limit of such subsequence is in every $K_i,$ showing that such intersection is not empty.  Thank you!
6862
Let $X$ be a metric space. Prove that if every continuous function $f: X \rightarrow \mathbb{R}$ is bounded, then $X$ is compact.  This has been asked before, but all the answers I have seen prove the contrapositive. Realistically, this may be the way to go, but is there way to exhibit an unbounded continuous function (under the assumption $X$ is not compact) without appealing to results beyond introductory real analysis (e.g., the solution I've seen involves the Tietze Extension theorem)?  Because we're working with metric spaces, it's clear that the assumption that $X$ is noncompact (towards proving contrapositive) will lead us to extract a sequence of points in the space with no convergent subsequences. But how can we infer the existence of an unbounded continuous on the space knowing only about some sequence of points in this space (without using anything too advanced)?
47578
Let $E_n$ be the set of all $f \in C\big([0,1]\big)$ for which there exists $x_0 \in [0,1]$ (depending on $f$) such that \begin{align*} \lvert\, f(x)-f(x_0)\rvert \leq n\lvert x-x_0\rvert, \end{align*} for all $x \in [0,1]$. Why is $E_n$ nowhere dense in $C\big([0,1]\big)$.  I have been able to show that $E_n$ is closed in $C\big([0,1]\big)$, but haven't been able to show why the interior is empty. Any advice?...
18101
Let $(X, d)$ be a metric space. I have to prove the following statements are equivalent.   $(X,d)$ is complete (i.e., every Cauchy sequence is convergent) and totally bounded (i.e., for every $\epsilon&gt;0$, $(X,d)$ has a finite $\epsilon$-net). $(X,d)$ is sequentially compact (i.e., every sequence has a convergent subsequence) . $(X,d)$ is Bolzano - Weierstrass compact (i.e., every infinite subset of $X$ has a limit point). $(X,d)$ is compact (i.e., every open cover of $X$ has a finite subcover). Every infinite open cover of $(X,d)$ has a proper subcover.   I have proved the following: 1$\Longleftrightarrow$2, 2$\Longleftrightarrow$3, 4$\implies$3, 2$\implies$4, 4$\implies$5 and 5$\implies$3. Although they are enough to prove the equivalence of all five statements, I tried to prove it from the other possible directions. But I'm stuck with the proof of 3$\implies$4 and 5$\implies$4. Does anybody know how to prove them without using any of the above statements? Please help.
50461
This could be classified as "homework", but I tried to solve this, made research online, and still failed, so I'll be glad to get some hints.  Let $G$ be a topological group, let $A$ be a compact subset of $G$, and let $B$ be a closed subset of $G$. Prove that $AB$ is closed.  If both $A$ and $B$ are not compact, but closed, this can fail, for example, if we let $A$ be the set of integers and $B$ the set of integer multiples of $\pi$, then both are closed, but $A+B$ is a proper dense subset of $\mathbb R$, so can't be closed. Also if $A$ is compact but $B$ is not closed, this easily fails.  Thanks
120092
Show that if $E$ is closed and $K$ is compact in $\mathbb{R}^n$, then $K+E$ is closed in $\mathbb{R}^n$, where $E+K=\{x+y\,\,\, x\in E \,\,\, , y \in K \}$ is Minkowski sum.  I tried to show it by saying that $K$ is closed and bounded but I do not know how proof goes?
92686
Fact  The intersection of a closed set $A$ and a compact set $B$ is closed.  Background  I wanted to prove that the intersection of closed and compact sets is compact. There might be other ways, but I am not interested in them for now and am looking to prove the above statement.  Relevant Facts  I don't have much idea what to use. $A^c$ is open. Every cover $\cup_{i\ge1}U_i$ of $B$ contains a subcover $\cup_{i=1}^kU_i$. I might want to prove $A\cap B^c$ is open. Limit points may be another points to consider but I don't get any idea in that direction.  I also don't have that $B$ is closed otherwise it would be trivial.  Some Clarification  I am aware that compact sets are not closed in arbitrary topological spaces. The reason why I was motivated to believe tha intersection of closed and compact set is compact is true is: See Matt E.'s Answer      So if $K$ is compact in an arbitrary topological space (which is just to say that   it is a compact topological space when given its induced topology) and     $L$ is closed then $K \cap L$ is a closed subset of $K$ in its induced topology   Is it true if $A$ and $B$ are not whole spaces?  I would appreciate any hints.