Let $X$ be a metric space , $d$ is the metric , show that $d$ is a continuous function from $X\times X$ to $R$.  I think the definition is all we need , but I just don't know where to start , can anyone help me.
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Given a metric space $(X,d)$, how to prove that the function $d \colon X \times X \to \mathbf{R}$ is continuous?   If we take any two arbitrary real numbers $a$ and $b$ such that $a &lt; b$, then we need to show that the set $d^{-1} (a,b)$ given by   $$ d^{-1} (a,b) :=  \{ (x, y) \in X \times X | a &lt; d(x,y) &lt; b \} $$  is open in the product topology on $X \times X$.   A basis for this product topology is the collection of all cartesian products of open balls in $(X,d)$.
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Suppose $(S,d)$ is a non-compact metric space. Is it possible to construct an unbounded continuous function from $S$ to $\mathbb{R}$? If it is possible, please show the construction method :)  Here's my attempt to solve this problem: Because $S$ is not compact, we can find a neighborhood assignment function $N(x)$ so that a finite cover is not possible. Then each time we can find an uncovered point from the union of neighborhoods before, and change the function by $f(x):=max\{f(x),\text{a function which equals to the indice of this point at that point and gradually vanishes at further points from the selected point}\}$. However I'm concerned that while this function is unbounded, there may be some invalid points for this function (goes to infinity)...  EDIT: As I found out there is a duplicate question, however the elementary solution is not presented in that question, so I still wish to know about this problem. :)
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My question: let $f : X \to \Bbb R$ be a non-constant continuous function on a connected metric space and assume that $f(X)$ is uncountable; then $X$ is uncountable.  We know continuous image of a connected metric space is connected. Since connected space in $\Bbb R$ are intervals and the function is non-constant continuous function so $f(X)$ is an interval in $\Bbb R$ and hence uncountable.  Is the proof correct?? How does it follows that $X$ is uncountable?
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Let $(X,d)$ be a metric space and $A\subset X$ and $x\in X$. Then  $x\to d(x,A)$ is a uniformly continuous function.  If $\partial A=\{x\in X\,:\,d(x,A)=0\}\cap\{x\in X\,:\,d(x,X-A)=0\}$, then $\partial A$ is closed for any $A\subset X$.  If $A,B$ are subsets of $X$  then $d(A, B)=d(B,A)$. If the function is uniformly continuous then $d(x,y)&lt;\delta $ implies $d(d(x,A)-d(y,A))&lt;\epsilon$ , I can not handle the last expression. Difficulty continues for 3rd choice also. At least give me some hints. And for the 2nd choice I think it is true, as it is intersection of two closed sets. As finite intersection of closed sets closed $\partial A$ closed. And logic for former two sets closed is that the sets are preimage of closed set (singleton $\{0\}$ is closed as $\mathbb R$ is $T_1$) under continuous map. I am not very much sure about my ideas and want a verification.
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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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Let $(X,d)$ be a metric space and $ a \in X $ Show that $f(x) = d(x,a)$ is a Lipschitz function from X to $\Re$ Use this fact to show $S$ is an open subset of $X$: Let $(X,d)$ be a metric space and $a \in X, r &gt; 0.$ Let $S$={$x \in X : d(a,x) &gt; r$}  &gt;Lipschitz Defn:  Let $f$ be a function from $(X,d)$ to $(Y,\rho)$ $f$ is said to be Lipschitz if: $\exists$ K $\geq$ 0 s.t. $\rho( f(x_1) , f(x_2) ) \leq$ Kd($x_1$,$x_2$) $\forall$ $x_1 x_2 \in X$  &gt;Continuity Defn:  Let $(X,d)$ and $(Y,\rho)$ be metric spaces, and $f: X \to Y$ Then $f$ is continuous $iff$ f$^{-1}$($B$) = { $x \in X : f(x) \in B $} is an open subset of $X$, whenever $B$ is an open subset of $Y$  &gt;Triangle Inequality for a Metric:  $d(x,z) \leq d(x,y) + d(y,z)$   My first question is, am I thinking about this correctly? If $f(x)$ is a Lipschitz function, it is uniformly continuous If $f(x)$ is uniformly continuous, it is continuous at every point If $f(x)$ is continuous at every point $\Rightarrow$ the inverse images of open sets are open Hence $S$ is an open subset of $X$.  Secondly, how would you formally construct this? We know $f(x) = d(x,a)$ s.t. $f:X \to \Re$ By definition of Lipschitz we have $\exists  K \geq 0$  s.t. $\rho(f(x_1),f(x_2)) \leq Kd(x_1,x_2) \forall x_1,x_2 \in X$ &amp; since our codomain is $\Re$ $\rho(f(x_1),f(x_2)) = | d(x_1,a) - d(x_2,a) | \leq Kd(x_1,x_2)$ Taking $K=1$ $\ldots$ Help in finishing this off would be appreciated. Thanks.
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If $(X,d)$ is a complete metric space and $A$ is closed in $(X,d)$ then show that for each $x \in X$ there exists an element $a_0 \in A$ such that $d(x,A)=d(x,a_0).$   I tried this problem several times but always got stuck at some point or another. For instance, I initially thought about showing the existence of a Cauchy sequence $(a_n)$ in A such that $d(x,a_n) \to \operatorname{inf}\{d(x,a):a \in A\}=d(x,A)$ and then by completeness of $A$ and by continuity of the function $f(a)=d(x,a)$, to prove the existence of an element $a_0$ in $A$ such that $a_n \to a_0$ which would imply $d(x,a_n)=f(a_n) \to f(a_0)=d(x,a_0)=d(x,A)$. But the problem was that I could not construct such a Cauchy sequence in $A$. All my other attempts have similarly gone in vain.  If anybody has any answer, please reply. Thank you.
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Let $(X,d)$ be a metric space and $S \subset X$. Show that $d_S(x):=\text{inf}\{d(x,s): s \in S\}=0 \Leftrightarrow x \in \overline S .$   Notes: $\overline S$ is the closure of S. Maybe you can use that a closed set is also closed for sequences in the set? I think the difficult part is when S is open, otherwise its trivial as the closure would be equal to S.
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Show that $(X,d)$ is a metric space where $X =\Bbb C $ and the distance function is defined as:   $$d(x,y) = \frac {2|x-y|}{\sqrt {1+|x|^2} +  \sqrt {1 + |y|^2}}, \text{ for } x,y \in \Bbb C.$$   I have done the proof of the first two propositions for being a metric, but I'm having a problem in proving the triangle inequality.
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Show that the bounded function $f(x) = \sin(x^2)$ is continuous, but not uniformly continuous, on the interval $[0,\infty)$.  The problem should be solved using the following definitions: Definition of Continuous Function Let $X$ and $Y$ be metric spaces. We sat that a function $f:X\rightarrow Y$ is continuous if given any open set $V \subseteq Y$, its inverse image $f^{-1}(V)$ is an open subset of $X$. Definition of Uniformly Continuous Function Let $(X,d_x)$ and $(Y,d_y)$ be metric spaces. If $E \subseteq X$, then we say that a function $f: X\rightarrow Y$ is uniformly continuous on $E$ if for every $\epsilon &gt; 0$ there exists a $\delta &gt; 0$ such that for all $x,y \in E$ we have $$d_x(x,y) &lt; \delta \longrightarrow d_y(f(x),f(y)) &lt; \epsilon$$ For the continuity part it is clear that $f^{-1}(a,b), (a,b) \subseteq (-1,1) = \{x:f(x) \in (a,b)\}$ which appears to be an open set however I doubt this is sufficient as a proof since it holds for any function $f$. Not sure how to go about proving $f$ not uniformly continuous. Seems to be difficult to prove from the definition alone.
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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'm trying to prove that $f:X\longrightarrow\mathbb{R}$ is continuous if, and only if, the sets $$\{x\in X:f(x)\geq\alpha\} \text{ and } \{x\in X:f(x)\leq\alpha\}$$ are closed $\forall\alpha\in\mathbb{R}$, where $X$ is a metric space.  I've managed to prove the $\left(\Rightarrow\right)$ direction:  I'm using the negative, that is, supposing that one of the sets are not closed and showing that $f$ won't be continuous. Suppose $A\equiv\{x\in X:f(x)\geq\alpha\}$ is not closed, then there's a limit point $y\notin \{x\in X:f(x)\geq\alpha\}$ and a sequence $y_n\longrightarrow y$. Without loss of generality, let $y_n\in A,\forall n\in\mathbb{N}$, then $f(y_n)\geq\alpha$, but $f(y)&lt;\alpha$.  Therefore, $|f(y)-f(y_n)|\geq\varepsilon &gt;0,\forall n\in\mathbb{N}$, where $\varepsilon = \alpha - f(y)&gt;0$. $\Box$  However, I'm having some trouble with the $\left(\Leftarrow\right)$. Can anyone give me some tips?  Thanks for helping! :D
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A function $f:D\subset \mathbb R \to \mathbb R$  is lipschitz given that there exists a $L\gt0$ such that $|f(x)-f(y)|\le L|x-y|$  I need to prove this function is then continuous. Is there a best definition of continuous functions to use for this proof given the definition of Lipschitz?  Is the epsilon delta way the only way to do it? can someone help me set it up? I really struggle with implementing that definition.
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I've been trying to proof that the metric space $(\mathbf{B}(T,X), d_\infty)$ is complete, if $(X,d)$ is complete. The definitions are shown below: \begin{align*}    \mathbf{B}(T,X) = \lbrace f: T \rightarrow X: f(T) \text{ is bounded} \rbrace \end{align*} and, \begin{align*}    d_\infty(f,g) = sup\lbrace d(f(t),g(t)):t\in T\rbrace \end{align*} I have proved that $d_\infty$ is a metric, but I'm stuck at proving that the metric space is complete :( Thanks :)
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Let $f$ be a function from a metric space $X$ to a metric space $Y$. Show that if $f$ is uniformly continuous on $X$ then $f$ is continuous on $X$. Show that the converse is not true.  Uniform continuity definition:  $∀ε&gt;0:∃δ&gt;0:∀p,q∈X:d_{X}(p,q)&lt;δ⟹d_{Y}(f(p),f(q))&lt;ε$  Continuity definition:  $∀ε&gt;0:∀p∈X:∃δ&gt;0:∀q∈X:d_{X}(p,q)&lt;δ⟹d_{Y}(f(p),f(q))&lt;ε$  I wonder if these proofs are correct and if they are formal enough?    Uniform continuity ⟹ Continuity   Let $f$ be uniformly continuous. Fix $ε_{0}$, obtain $δ_{0}(ε_{0})$ (as a function of $ε_{0}$), fix any $p_{0}$ and $q_{0}$ and we know that:   $d_{X}(p_{0},q_{0})&lt;δ_{0}⟹d_{Y}(f(p_{0}),f(q_{0}))&lt;ε_{0}$   If we want to prove that $f$ is continuous at $p_{0}$, we fix $ε_{0}$ and we pick the same $δ_{0}$ as above and fix any $q_{0}$ and we are assured by 1. that   $d_{X}(p_{0},q_{0})&lt;δ_{0}⟹d_{Y}(f(p_{0}),f(q_{0}))&lt;ε_{0}$   Since $p_{0}$ was arbitrary, $f$ is continuous at $X$.     Continuity $\color{red}{\implies}$ Uniform continuity   Let $f$ be continuous. Fix $ε_{0}$, fix any obtain $p_{0}$ and $p_{1}$, obtain both $δ_{0}(ε_{0},p_{0})$ and $δ_{1}(ε_{0},p_{1})$ (not necessarily equal), fix any $q_{0}$ and $q_{1}$ and we know that:   $d_{X}(p_{0},q_{0})&lt;δ_{0}⟹d_{Y}(f(p_{0}),f(q_{0}))&lt;ε_{0}$  $d_{X}(p_{1},q_{1})&lt;δ_{1}⟹d_{Y}(f(p_{1}),f(q_{1}))&lt;ε_{0}$   For $f$ to be uniformly continuous, if we fix $ε_{0}$, obtain $δ$, fix any $p_{0}$, $p_{1}$, $q_{0}$ and $q_{0}$ we must have that:   $d_{X}(p_{0},q_{0})&lt;δ⟹d_{Y}(f(p_{0}),f(q_{0}))&lt;ε_{0}$  $d_{X}(p_{1},q_{1})&lt;δ⟹d_{Y}(f(p_{1}),f(q_{1}))&lt;ε_{0}$   But those two implications need not hold for the same $δ$ as shown above. Suppose we pick $δ_{0}$ and w.l.o.g. we suppose $δ_{0}&gt;δ_{1}$. Then it could be the case that   $δ_{1}&lt;d_{X}(p_{1},q_{1})&lt;δ_{0}$  but we cannot conclude ⟹$d_{Y}(f(p_{1}),f(q_{1}))&lt;ε_{0}$.  Therefore, continuity does not imply uniform continuity.
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Let $f$ denote a map from a product space $X \times Y$ to $Z$. If for every $x\in X$, the map $f(x,-)$ is continuous, and the same holds for every $y \in Y$, then is $f$ continuous in general? If not, is there any condition to be imposed to make $f$ continuous?
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Working on the following problem from Munkres:      Let $(X, d_{X})$ and $(Y, d_Y)$ be metric spaces; let $Y$ be complete.   Let $A \subset X$. Show that if $f:A \to Y$ is uniformly continuous,   then $f$ can be uniquely extended t a continuous function $g:  \overline{A} \to Y$, and $g$ is uniformly continuous.   If there exists such an extension, it must be unique, and $f(x) = \lim_{n \to \infty}f(x_n)$ for $x_n$ a sequence in $A$ converging to $x \in \overline{A}$. To show this limit exists, let $x_n \to x$, then $f(x_n)$ is Cauchy, since uniformly continuous functions take Cauchy sequences to Cauchy sequences. Then $f(x_n)$ converges by completeness, and I claim it converges to $f(x)$.  Now, to show $g$ is uniformly continuous, I kind of brute-forced it, and I was wondering if anyone had a more elegant method. (I consider the proof up to now pretty elegant.) What I did was: Suppose for contradiction that $\exists \epsilon &gt;0$ s.t. $\forall \delta &gt;0\,\, \exists x_\delta, y_\delta \in \overline{A}$ s.t. $d(x_\delta, y_\delta)&lt; \delta$ and $d(g(x_\delta), g(y_\delta))&gt; \epsilon$. Let $\delta := \delta_{\epsilon}/2$, where $\delta_\epsilon$ is taken from the uniform continuity of $f$. Now $d(g(\cdot), g(\cdot))$ is continuous...in particular, it is continuous in each variable separately. Therefore we can vary $x_{\delta}$ and $y_{\delta}$ a little bit, to get $x^*,\, y^* \in A$ such that $d(x^*, y^*)&lt;\delta_{\epsilon}$ and $d(f(x^*), f(y^*))&gt;\epsilon$, a contradiction.   Does anyone know a more elegant method to show the uniform continuity?