Why is a continuous mapping from a compact metric space to another metric space is uniformly continuous? This theorem is from Rudin Real Analysis Page 202.
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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.
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A contractive mapping on $M$ is a function $f$ from the metric space $(M,d)$ into itself satisfying $$d(f(x),f(y))&lt;d(x,y)$$ whenever $x,y\in M$ with $x\ne y$. Prove that if $f$ is a contractive mapping on a compact metric space $M$, there exists a unique point $x\in M$ with $f(x)=x$.   Consider the function $g(x)=d(f(x),x)$. Note that   \begin{align*}d(f(x),x)-d(f(y),y)&amp;\leq (d(x,y)+d(y,f(x))-(d(y,f(x)-d(f(x),f(y)) \\ &amp;=d(x,y)+d(f(x),f(y)) \\ &amp;&lt;2d(x,y)  \end{align*}  And similarly for $d(f(y),y)-d(f(x),x)$, so we have $|d(f(x),x)-d(f(y),y)|&lt;2d(x,y)$. This means $g(x)$ is continuous.  So $g(x)$ is a real-valued continuous function on a compact space, so must be bounded and has a minimum $c$, i.e. there exists $x_0$ such that $d(f(x_0),x_0)$ equals the minimum $c$. What then?
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Let $f$ be a continuous function. If $f$ is uniformly continuous then it maps cauhy sequence into cauchy sequence. Is the converse true? That is if $f$ maps cauchy into cauchy gives the uniform continuity of $f$. I think it's no. Can suggest a counter example
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Let $X$ and $Y$ metric spaces, $f$ is an injective from $X$ to $Y$, and $f$ sets every compact set in $X$ to compact set in $Y$. How to prove $f$ is continuous map?  Any comments and advice will be appreciated.
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Let f and g be continuous mappings of a metric space $X$ into a metric space $Y$ and let $E$ be a dense subset of $X$. Prove that $f(E)$ is dense in $f(X)$. If $g(p)=f(p)$ for all $p \in E$, prove that $g(p)=f(p)$ for all $p \in X$.
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Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. $f_n: X\to Y$ with $n\in\mathbb{N}$ and $f:X\to Y$ functions. $f_n$ is continuous for every $n$ and $f_n\stackrel{n\to\infty}{\longrightarrow} f$ uniformly.       Then $f$ is continuous.   Here is my proof:  Let $x\in X$ and $\epsilon &gt;0$ be arbitrary.  Since $f_n\to f$ uniformly it exists $N\in\mathbb{N}$ such that $d_Y(f_N(x), f(x))&lt;\epsilon/3$ for every $x\in X$.  As $f_N$ is continuous we have for $x_0\in X$ and $\delta &gt; 0$, that $d_Y(f_N(x),f_N(x_0))&lt;\epsilon/3$ if $d_X(x,x_0)&lt;\delta$.  This gives us:  $$\begin{align}d_Y(f(x),f(x_0)&amp;\leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f(x_0))\\ &amp;\leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f_N(x_0))+d_Y(f_N(x_0), f(x_0))\\ &amp;&lt; \epsilon/3+\epsilon/3+\epsilon/3=\epsilon\end{align}$$  Thanks in advance for your correction.
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I understand the formal definition of uniform continuity of a function, and how it is different from standard continuity.  My question is: Is there an intuitive way to classify a function on $\mathbb{R}$ as being uniformly continuous, just like there is for regular continuity? I mean, that for a "nice" function $f:\mathbb{R} \to \mathbb{R}$, it is usually easy to tell if it is continuous on an interval by looking at or thinking of the graph of the function on the interval, or on all of $\mathbb{R}$. Can I do the same for uniform continuity? Can I tell that a function is uniformly continuous just by what it looks like? Ideally, this intuition would fit with the Heine-Cantor theorem for compact sets on $\mathbb{R}$.
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This is the analogue of closed graph theorem for compact space          Suppose that $X$ and $K$ are metric spaces, that $K$ is compact, and that the graph of $f: X \rightarrow K$ is a closed subset of $X \times K$. Prove that $f$ is continuous.      Because metric space is first countable, all we need is to prove that if $x = \lim_{n \rightarrow \infty} x_{n}$, then $f(x) = \lim_{n \rightarrow \infty} f(x_{n})$. Suppose $x = \lim_{n \rightarrow \infty} x_{n}$, then because $K$ is compact, sequence $\{f(x_{n})\}$ has a subsequence $\{f(x_{n_{i}})\}$ converge. Because the graph of $f$ is closed, we must have $\lim_{n \rightarrow \infty}\{f(x_{n_{i}})\} = f(x)$. All we need to prove now is $\{f(x_{n})\}$ converge to $f(x)$, but I can't prove that. Can any one help me. I really appreciate. Thanks
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This is an exercise from Thomson and Bruckner's "Elementary Real Analysis" (Exercise 5.6.5 page 239):     Let $X=\{x_1,x_2,\dots,x_n,\dots\}$. What property must $X$ have so that every function continuous on $X$ is uniformly continuous on $X$?   The set $X$ is a subset of $\mathbb {R}$. I worked on the $\varepsilon-\delta$ definitions of limits and uniform continuity, but I coudln't link between them. Could you please help me?  Edit: I removed the part My attempt because my result was wrong.
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Suppose that $f:\Bbb R\to \Bbb R$ is continuous and that $f(x)\to 0$ as $x\to \pm\infty$. Prove that $f$ is uniformly continuous.   Since $f(x)\to 0$ as $x\to \pm\infty$ we have $|f(x)|&lt;\epsilon $ whenever $x&lt;-K$ and $x&gt;K$ where $K&gt;0$.  Also $f$ is uniformly continuous on $[-K,K]$ as the domain is compact.  Hence $f$ is uniformly continuous in $(-\infty,-K)\cup [-K,K]\cup (K,\infty)$.  But how to conclude that $f$ is uniformly continuous on $\Bbb R$ from above as I may chose $x\in (-\infty,-K);y\in [-K,K]$ ,how to conclude that $|f(x)-f(y)|&lt;\epsilon $ from above for a given $\epsilon$?  Please suggest the edits required.
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Show $f:(0,\infty)\rightarrow \mathbb{R}, f(x)=x\ln x$ on $(0,\infty)$ is not uniformly continuous.    I think that the general way to prove that something is not continuous in a metric space is to let $\epsilon=...$ and show that $\forall\delta&gt;0$, $d&#39;(f(x)-f(y))&gt;\epsilon$. I can't use the Mean Value Theorem because we haven't gone over it yet.    Here's an attempt: Let $\epsilon=1$. Without loss of generality, let $x&gt;y&gt;0$. $|x\ln x-y\ln y|&lt;|(x-y)(\ln x)|\leq|x-y||\ln x|$...$\delta=\frac{\epsilon}{\ln x}$ and therefore doesn't work for all $x$?
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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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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?
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Here's a question I came up randomly just now:  If a sequence of absolutely continuous functions $\{\phi_{n}\}$ converges uniformly to a function $f$, does it imply that $f$ is also absolutely continuous?  I am having trouble judging if this is true. If the statement is not true, can anybody help me find a counter-example?  Thank you!
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It is known that if $f(x):R\to R$ is absolutely continuous on $[a,b]$, then it has bouded variation on $[a,b]$. Is it also true for uniformly continuous functions? And if not then what is an example?
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Whenever $X$ is a topological space, let us say that continuity on $X$ is detected by $\mathbb{R}$ iff for all functions $f : X \rightarrow Y$ where $Y$ is another topological space, we have that if every continuous function $c : \mathbb{R} \rightarrow X$ satisfies "$f \circ c$ is continuous", then $f$ is itself continuous.   Clearly, continuity on $\mathbb{R}$ is detected by $\mathbb{R}$ It is shown  that continuity on $\mathbb{R}^2$ is detected by $\mathbb{R}$. Clearly, it is not the case that continuity on $\mathbb{Q}$ is detected by $\mathbb{R}$, since every continuous mapping $\mathbb{R} \rightarrow \mathbb{Q}$ is constant.      Question. Do there exist general conditions underwhich we can conclude that continuity on a topological space is detected by $\mathbb{R}$?
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I am thinking about this since couple hour. Is a continuous and bounded function $f:\mathbb R\to\mathbb R$ uniform continuous too? I didn't found a counter example and thus I tried to prove this like this: let $\epsilon&gt;0$ and $a\in\mathbb R$. By continuity, there is a $\delta_a&gt;0$ such that $$d(f(x),f(y))&lt;d(f(x),f(a))+d(f(y),f(a))&lt;\epsilon$$ if $d(x,y)\leq d(x,a)+d(y,a)&lt;\delta_a.$ Then, if $\inf_a\delta_a&gt;0$ I set $\delta=\inf_a\delta_a$ and we are done since $d(f(x),f(y))&lt;\epsilon$ if $d(x,y)&lt;\delta,$ but I didn't success to prove that $\delta\neq 0$.