Let $M$ be a metric space. If every continuous function $f : M \rightarrow \mathbb{R}$ has compact range then $M$ is compact.  I got this question from "Real Mathematical Analysis" by Charles Chapman Pugh. Specifically Exercise 118 in Chapter 2. I'm not sure how to go about proving this question, but it seems useful for another equivalence of what it means for a metric space to be compact.  EDIT: I don't see how this question is a duplicate of the other link in question. Or I at least don't understand how it is a duplicate.
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Suppose $(M,d)$ a metric space. I want to show that      If every continuous real-valued function on $M$ attains a maximum, then $(M, d)$ is compact.   I was trying to do this by assuming $M$ non-compact and showing explicitly a function which does not attain a maximum. If $M$ is not bounded it's enough to take the distance from a fixed point as we can always find a sequence which "exits" every compact and pushes the distance at infinity. But what if $M$ is bounded?
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The title is a claim my classmate made during our summer vacation :D  He showed me a TeX file describing a proof of his claim, and it contains a fairly short but elegant proof. He says that the motivation was to see if the converse of the statement '$(X, d)$ is compact $\Rightarrow$ All metrics homeomorphic to $X$ is bounded.  I was surprised by his work, and wanted to know if it's already discovered and/or published.  So the question is,    Is the claim my classmate made is already discovered? If it is, any references?  How about more general one? like, can we remove/reduce the condition of the claim more weaker?  Are there any underlying theorems or theories in it? If it is not yet discovered, where should I recommend him to publish his work? (LOL) I think it is worth sharing the proof.
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Let $X$ denote a metric space.  Whenever $S \subseteq X$ and $r \in \mathbb{R}_{\geq 0}$, write $f(S,r)$ for the following set.  $$\{x \in X \mid \exists s \in S : d(x,s) \leq r\}$$     Question. Given a closed set $S \subseteq X$ and any $r \in \mathbb{R}_{\geq 0}$, does it follow that $f(S,r)$ is closed?   If not, a counterexample $(X,S,r)$ would be appreciated.
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Claim: Let $X$ be a metric space. If $A,B\in X$ are disjoint, if A is compact, and if B is closed, then $\exists \delta&gt;0: |\alpha-\beta|\geq\delta\;\;\;\forall\alpha\in A,\beta\in B$.  Proof. Assume the contrary. Let $\alpha_n\in A,\beta_n\in B$ be chosen such that $|\alpha_n-\beta_n|\rightarrow0$ as $n\rightarrow \infty$.  Since A is compact, there exists a convergent subsequence of $\alpha_n\;(n\in\mathbb{N})$, $\alpha_{n_m}\;(m\in\mathbb{N})$, which converges to $\alpha\in A$.  We have   $$|\alpha-\beta_{n_m}|\leq|\alpha-\alpha_{n_m}|+|\alpha_{n_m}-\beta_{n_m}|\rightarrow0 \;\;\;as\;\;m\rightarrow\infty.$$ Hence $\alpha$ is a limit point of B and since B is closed $\alpha\in B$, contradiction.  Is my proof correct? I feel as though I am missing something simple which would trivialize the proof.
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I'm a beginner in metric space. So many books I've read, there is only the notion of open covers. I want to know why do we worry about open covers to define the compactness of metric spaces and why don't we use closed covers? What is the problem in defining closed cover of a set? Can we use the alternative definition of compactness: "Every closed cover has a finite subcover"?
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Let $(X, d)$ be a compact metric space, and suppose $f : X → X$ satisfies    $$d(f(x), f(y)) &lt; d(x, y)$$   for all $x \neq y \in X$. Show that f has a unique fixed point.   All I've gotten it so far is that we need to somehow use another function $g(x)=(x,f(x))$.  Thanks
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Is it true, that if $X$ is a separable metric space, then the space of all continuous functions on $X$ with the supremum metric is also separable?
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Let $X$ and $Y$ be metric spaces, $X$ compact, and $T:X \to Y$ bijective and continuous. Show that $T$ is a homeomorphism.   My attempt:  We need only show that $T^{-1}$ is continuous.  Let $M \subset Y$ be closed. Then, from the surjectiveness of $T$, there exists a $A \subset X$ such that $T(A)=M$. Since $M$ is closed, by the continuity of $T$, we must have that $A$ is closed.  Now $$T^{-1}(T(A))=T^{-1}(M)$$ i.e, $A=T^{-1}(M)$. Since $A$ and $M$ are closed, we must have that $T^{-1}$ is continuous.  My problem with this, however, is that I am not using the fact that $X$ is compact anywhere in the proof. Is there perhaps another way that I can show this, using the compactness of $X$?   Any hints will be appreciated.
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Let $f : (X, d_X) \to (Y, d_Y )$ be a map between two metric spaces. Recall that $f$ is called continuous at $x ∈ X$ if for any open ball $ B_f(x)(ϵ)$ of radius $ϵ$ around $f(x)$ there exists a ball $B_x(δ)$ of radius $δ$ around $x$ such that $f(B_x(δ))⊂B_f(x)(ϵ)$. Show that $f$ is continuous at $x$ if and only if for any sequence ${x_n}$ in $X$ converging to $x$ the sequence $f(x_n)$ converges in $Y$ to $f(x)$. Hint: imitate the proof of the corresponding statement for functions $R → R$.
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The question is (from Topology without tears) that: Let $(X,d)$ be a metric space and $\tau$ the corresponding topology on $X$. Fix $a \in X$. Prove that the map $f:(X,\tau) \rightarrow \mathbb{R}$ defined by $f(x) = d(a,x)$ is continuous.  My first attempt was that let an open ball, $B_{\epsilon}(a)\subset \mathbb{R}$ and show that $f^{-1}(B_{\epsilon}(a)) \subset \tau$. Since $f(x) = d(a,x)$, $f^{-1}(B_{\epsilon}(a))=B_{\epsilon}(a) \in \tau$.   I am pretty sure the last sentence is wrong. I am not sure if every open ball is an open set in topology induced by metric space.  Please give me some hint and direction. Thanks
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This might be a duplicate but I tried googling the MSE site and could not find a satisfactory answer.  Let $(X, d)$ be a metric space and $f$ be a real valued continuous function on $X$. Suppose $f$ has a compact support. Does this imply the uniform continuity of $f$?   I tried proving this statement,  but only for locally connected spaces have I succeeded in doing so. Is thus true for general metric spaces? I just couldn't provide a proof (or a counterexample) by myself. Please enlighten me.
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I just got my exam back, and I still cannot understand this question:     Given a continuous function $f:A\subseteq\mathbb{R}\to\mathbb{R}$, show that if $A$ is a compact set, then its image, $f(A)$, is also compact.   I know that a set $A\subseteq\mathbb{R}$ is compact if every sequence in $A$ has a subsequence that converges to a limit that is also in $A$, and I know that a function $f$ is continuous on $A$ if for every $(x_n)\subseteq A$ such that $x_n\to c\in A$, it follows that $f(x_n)\to f(c)$. Therefore, all that I need to do is show that for every $(y_n)\subseteq f(A)$, there is a subsequence $(y_{n_k})$ such that $y_{n_k}\to y\in f(A)$.  Can I then make the assumption that for any sequence $(y_n)\subseteq f(A)$, there is a sequence $(x_n)\subseteq A$ such that $y_n=f(x_n)$? If so, I could then continue by stating that since $A$ is compact, there is a subsequence $(x_{n_k})$ such that $x_{n_k}\to x\in A$, and since $f$ is continuous, $f(x_{n_k})\to f(x)$. I believe that this yields the required subsequence $(y_{n_k})$ of $(y_n)$ such that $y_{n_k}=f(x_{n_k})\to f(x)=y\in f(A)$.  What do you guys think? Is this a sound approach? Thanks in advance.
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As the title indicates, I'd like to prove the following:     If $f:\mathbb R\to\mathbb R$ is a continuous function on $[a,b]$, then $f$ attains its maximum.   Now, I do have a working proof: $[a,b]$ is a connected, compact space, which means that because $f$ is continuous, $f([a,b])$ is compact and connected as well.  Therefore, $f([a,b])$ is a closed interval, which means it has both a minimum and, as desired, a maximum.  What I would like, however, is a proof that doesn't require such general or sophisticated framework.  In particular, I'd like to know if there's a proof that is understandable to somebody beginning calculus, one that (at the very least) doesn't invoke compactness.  Any comments, hints, or solutions are welcome and apreciated.
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let $ f_n :X \rightarrow R $ be sequence of continuous function on complete metric space $(X,d)$, which convergence pointwisely to $ f: X\rightarrow R  $ mean that $ f_n(x)\rightarrow f(x)$ $ \forall_x \in X$ I have to prove that set of continuity points is dense in $(X,d)$. I did it using Baire theorem, but I also have to prove that assumption of complete metric space is necessary. There is a hint that I should use $(Q,d_e)$ as a metric space to find a counterexample. So I started to think about a function which is uncontinuous in every point. I set $Q$ in a string  $q_{n}$ then take $f(q_{n})=1/n$ and as a function $f_{m}$ take any continuous function(I was thinking about linear, but how to create such?)  such that  $ f_{m}(q_{n})=1/n$ for  $n=1,2,...,m$ Then $f_{m}$ converges pointwisely to $f$. Any hints? Am I thinking correctly?
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The graph of $f$ is $G(f) = \{(x,f(x)) : x\in X\} \subseteq X\times Y$  $X$ and $Y$ are metric spaces.  a) Suppose $f$ is continuous and prove that $G(f)$ is a closed set.  b) Suppose that $G(f)$ is compact and prove that $f$ is continous  For a), the definition of a closed set that comes to my mind is a set that contains all its limit points (or was it accumulation points?), is there another equivalent definition that may b more helpful to prove a)? Is it possible to prove this directly? Because at first glance the only way I could imagine to prove this is by contradiction or contrapositive.   I imagine that the proof of b) will be immediately derived from a).
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If $(X,d)$ is limit point compact, show it is compact.  I have found multiple proofs of this that first show that limit point compact implies sequential compact, which in turn implies compactness. I was wondering if there is a direct proof showing that limit point compactness implies compactness in a metric space.
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I am struggling with this question:     Prove or give a counterexample: If $f : X \to Y$ is a continuous mapping from a compact metric space $X$, then $f$ is uniformly continuous on $X$.    Thanks for your help in advance.