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.

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Proof that a function with a discontinuity has Darboux Property

I have the following function: $$f:\Bbb R\to \Bbb R\text{ defined by } f(x)=\left\{\begin{matrix}0 & \text{ if } x\leq0, \\ \cos(x)\sin\left(\frac1{x}\right) &\text{ if } x>0. \end{matrix}\right. $$ I have to prove that this function has the Darboux…
Wolfuryo
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can you find the geometric mean of a continuous variable, in a similar why to find the mean of a continuous variable.

Can you find the geometric mean of a continuous variable, in a similar why to find the mean of a continuous variable. To find the regular mean of between to points on a function you do $\frac{\int_{a}^{b} f(x) \,dx}{b-a} $ If there is 1 point that…
user913631
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A value problem on continuous function

Let $f\colon[1,8]\to \Bbb{R}$ be a continuous function such that $f(1)=4$ and $f(8)=3$. Show that exists $x_0 \in [1,8]$ such that $$f(x_0)= \sqrt[3]{x_0}+x_0.$$ How can I start this?
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How can one prove that a function is continuous?

I have the given function, $$ f(x,y)=1+\sin^2(xy)+\cos(x+y) $$ I am wondering how can I prove that it is continuous on its domain? The limit of this function is 2 as x,y approach zero. However I am not sure how to use the limit to prove…
user287546
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local extrema of a Continuous function

Let be a continuous function $f\colon \mathbb{R}\to \mathbb{R}\:$ that has exact 3 local extrema. $f$ is NOT differentiable. Find the maximum number of local extrema that the function $f\circ f$ can have. I do not know how to prove that but I…
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Continuity in a piecewise function defined separately on rational and irrational numbers

A question defines an f(x) that is x when x is rational and 1-x when x is irrational, and asks for the points where the function is continuous. The answer equates the two expressions and says the answer's $\frac{1}{2}$. Why is this so? I can only…
harry
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If $f(x) = x^3 - x^2 + (x-1)\sin x$, then which of the following statements is/are true?

$f(x) = x^3 - x^2 + (x-1)\sin x$ $fg(x) = f(x) g(x)$ Which of the following is true? A) If g is continuous at x = 1, then fg is differentiable at x = 1 B) If fg is differentiable at x = 1, then g is continuous at x = 1 C) If g is differentiable at x…
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Show that $f^{-1}: S^{1} \to [0,2\pi)$ is not continuous

I have $f^{-1}: S^{1} \to [0,2\pi)$ where $(cos(t), sin(t)) \mapsto t$. I want to show that this function is not continuous by using the epsilon-delta defintion $\forall\epsilon >0 \, \exists\delta>0 \,\, \forall t\in S^{1}: |t-t_{0}|<\delta…
user804292
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Showing a continuous function between two topologies is constant

Let $X$ be an arbitrary set, and let $f: \Bbb R → X$ be a function that is continuous with respect to the euclidean topology on $\Bbb R$ and the discrete topology on $X$. Prove that $f$ is constant. So basically, in a different scenario, we're given…
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Neighbourhood property of continuity

Let $c\in\Bbb R$ and a function $f:\Bbb R\to\Bbb R$ is continuous at c. If for every positive $\delta$ there is a point $y\in(c-\delta, c+\delta)$ such that $f(y)=0$, prove that $f(c)=0$.
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Continuity of $f(x)= \sum _ {n=0} ^{\infty } \left ( \frac{x^n}{n!} \right )^2$

Recently, I've been asked to prove that this function is continuous on $\mathbb{R}$ $$f(x)= \sum _ {n=0} ^{\infty } \left ( \frac{x^n}{n!} \right )^2$$ I do recognize its very similar to the numerical series that defines $e$, yet I am working here…
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Is the image of a dense set of a bijective continuous set dense?

I have a dense set $K\subset (0,\pi)$. We have the continuous bijective function $h=\theta\mapsto\cot(\theta)$. This function $h:(0,\pi)\rightarrow \mathbb{R}$. My question is: is $h(K)$ dense in $\mathbb{R}$? The main problem I have with this…
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Continuity of a function at a point in $\mathbb{R}_+$

I am given a function where $$f:S\subset \mathbb{R}_+\rightarrow \mathbb{R} $$ $$f(x)=\frac{\sqrt{x}-1}{x-1}$$ I need to use the following definition to prove that $f$ is continuous at $z\in S$; $f(x)$ converges to the limit, L, as $x$ tends to $z$,…
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Is $f(y) = \frac{\sin y}{y}$ a continuous function?

Is $f(y) = \frac{\sin y}{y}$ a continuous function? I am not sure about the point at $y=0$, the denominator cannot be zero but the numerator is also zero! The limit at $y=0$ exists but what about the original function?
Smarties
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Continuity of distribution function

Given $$F(x)= \begin{cases} 0 &x<0\\ x &0\leq x\leq1\\ 1 &x>1\end{cases}$$ Why this distribution function is discontinuous at $x=0$ and $x=1$ But according to me $F(1)=1$ and $$\lim(1-h)=1 F(1+)=1$$ and similarly for $0$…