Questions tagged [continuity]

Intuitively, a continuous function is one where small changes of input result in correspondingly small changes of output. Use this tag for questions involving this concept. As there are many mathematical formalizations of continuity, please also use an appropriate subject tag such as (real-analysis) or (general-topology)

Analytic Definition: Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces. A map $f : X \to Y$ is said to be continuous at $x_0$ if for every $\varepsilon > 0$, there is $\delta > 0$ such that $d_Y(f(x_0), f(x)) < \delta$ whenever $d_X(x_0, x) < \varepsilon$. A map $f : X \to Y$ is said to be continuous if it is continuous at $x_0$ for all $x_0 \in X$.

Topological Definition: Let $(X,\mathcal T_X)$ and $(Y, \mathcal T_Y)$ be topological spaces. A map $f : X \to Y$ is said to be continuous if $U \in \mathcal T_Y$ implies that $f^{-1}(U) \in \mathcal T_X$.

In the case of metric spaces, the metric induces a topology, and the two notions of continuity coincide. Note that multiple metrics can induce the same topology, and that not all topologies are metrizable (can be generated from some metric).

Continuity is a sufficient condition for the intermediate value theorem. It is also a necessary condition for the extreme value theorem, as well as differentiability.

17100 questions
1
vote
0 answers

In this link I am not getting why is An is finite?

In this link, I am not getting why is $A_n$ is finite? Prove that function is continuous in all irrational points Why is the number of such positive integers at most $[\frac {1} {\epsilon}]$ , the integral part of ${1\over\epsilon}$? And why is…
1
vote
3 answers

Continuity of $f(x,y) = \dfrac{\sin(x^2+y^2)}{x^2+y^2}$ at $(x,y) = (0,0)$

Continuity of $f(x,y) = \dfrac{\sin(x^2+y^2)}{x^2+y^2}$ at $(x,y) = (0,0)$ I need to specify the value of $f(x,y)$ which makes the given function continuous. I had tried to give a arbitral relationship between $y$ and $x$, but as $(x, y)$ gets…
Daschin
  • 675
1
vote
2 answers

continuity, uniformly continuous

I was wondering if you could help me with this: $$ f(x)=\begin{cases} x^2\sin\left(\frac{1}{x}\right) & \text{ for } x \ne 0\\ 0 & \text{ for } x=0 \end{cases}. $$ I need to observe that f is continuous on $\mathbb{R}$ and then explain why it is…
ulc8
  • 45
1
vote
1 answer

Continuity of the distance function

Let $A$ be convex compact subset of $\mathbb{R}^n$. For each $x\in \mathbb{R}^n$, there exists $y_x\in A$ such that $d(x,y_x)=d(x,A)$. Defined function $P_A:\mathbb{R}^n\to A$ where $P_A(x)=y_x$. Prove that $P_A$ is continuous?. Here $d$ denote…
Muniain
  • 1,453
1
vote
1 answer

Continuity Values

I have been trying to figure this question out for a while now, it's in a past paper I'm studying for my Calculus exam tomorrow. But we don't get given the memo's. I have done simpler versions of this question where it was a straight line, but this…
Nick Corin
  • 251
  • 3
  • 9
1
vote
2 answers

$f:(0,1)\rightarrow \mathbb{R}$ and $f(x)= x^{2} $ if $x$ is rational and $2-x^{2}$ if x is irrational. Where is f continuous?

Is this function not Continuous anywhere in $(0,1)$, because the value of the function will keep oscillating between $ x^{2} $(for rational) and $2-x^{2}$ (for irrational) values, So there will be breaks in the function??
1
vote
1 answer

Cantor set: discontinuities of its characteristic function

Let C denote the Cantor set constructed by removing the typical open middle thirds inductively from the interval [0,1] in R. I denote as f the characteristic function of C, i.e., f(x)=1 if x belongs to C and f(x)=0 otherwise. I guess that the set of…
1
vote
3 answers

Proving continuity using $\delta $- $\varepsilon$

The question is: Given the function $f:[-1,1]\to\Bbb R$, defined by $$f(x)= \begin{cases}x\sin\frac{1}{x},& \text{for }x\neq 0\\ 0,&\text{for }x=0. \end{cases}$$ Prove that the function admits maximum and minimum in its domain (Note that you are not…
FJA
  • 11
1
vote
0 answers

Proof of parallel chord theorem

Where can i find a correct proof of the following theorem of H.Hopf: For every continuous curve $L$ in the plane with endpoints $A$ and $B$ such that distance $\vert AB \vert = 1$, and for any natural number $n$, there exists a chord (meaning a…
exp8j
  • 587
1
vote
3 answers

Is the function $f( x)=1/|x|^{1/2}$ Lipschitz continuous?

Is the function $f( x)=1/|x|^{1/2}$ Lipschitz continuous near $0$? If yes, find a constant for some interval containing $0$ I think the answer is yes since I can find $L=1$ that satisfies Lipschitz continuity criteria in a interval close to zero,am…
Klara
  • 1,233
1
vote
3 answers

Is $\frac{x-|x|}{x}$ continuous?

Is $\frac{x-|x|}{x}$ continuous? It should be discontinuous at x=0 as left hand limit and right hand limit at zero are unequal. But it is continuous and the reason given is that they have excluded zero as its domain. Is it possible?
1
vote
2 answers

Find the value of p for the functionto be continous

In the following function . We have to determine the value of p , if possible, so that the function is continuous at $x = 1/2$. I tried and found the LHS of function at $x=1/2$ that is $(-1/4)$. But how could we comment on RHS
1
vote
1 answer

Functions and continuity.

Let $f : \mathbb{R} \to \mathbb{R}$ be continuous at $c$ and let $f(c) > 0$. Show that there exists an open interval $I$ containing $c$ such that $f(x) > 0$ for all $x \in I$.
1
vote
2 answers

Suppose a function f which maps a real no to a real no is an odd function satisfying $\lim_{x\to 0^+} f(x) = f(0)$

Suppose a function f which maps a real no to a real no is an odd function satisfying $\lim_{x\to 0^+} f(x) = f(0)$ Show that $f(0)=0$ and f is continuous at $x=0$ I am stuck at proving the continuity part. Heres my partial solution Since f is an…
Yellow Skies
  • 1,710
1
vote
2 answers

If f is continuous function.Then inverse image of closed set is closed.

I know the above quoted statement is correct but I have a confusion in one example.... f(x)=sin x f: (π/4 , 7π/4)--> [-1,1]…… If we take the closed set in the range as the whole set [-1,1] then its inverse image will be the whole…