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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I think I found a flaw in the $\varepsilon$-$\delta$ definition of continuity.

If I have a function $f(x)$ defined as follows. $f(x) = 1$ for all $x<1$ and $x>2$; $f(x) = 100$ for $x = 1.5$; $f(x)$ is undefined anywhere else. According to the $\varepsilon$-$\delta$ definition of continuity, if I take $\delta$ as any positive…
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Can a cube of discontinuous function be continuous?

Can a cube (meaning $g(x) = f(x)^3 = f(x) \cdot f(x) \cdot f(x)$) of discontinuous function $f: D \to \mathbb{R}$ ($D$ is subset of $\mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a…
J. Abraham
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Continuity on closed intervals - differentiability on open intervals

In our lectures notes, continuous functions are always defined on closed intervals, and differentiable functions, always on open intervals. For instance, if we want to prove a property of a continuous function, it would go as "Let $f$ be a…
user62487
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Does a function have to be "continuous" at a point to be "defined" at the point?

I did search for whether this question was already answered but couldn't find any. Does a function have to be "continuous" at a point to be "defined" at the point? For example take the simple function $f(x) = {1 \over x}$; obviously it is not…
jamadagni
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Can a function be considered continuous if it reaches infinity at one point?

Say we have a function $ f = \dfrac{1}{\arctan|x|^3} $ If we add to that definition with $ f(0) = +\infty $ Can $ f$ now be considered continuous? I'm assuming you can't just say that function equals infinity at one point. If we can't do that, is…
Luka Horvat
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does the uniform continuity of $f$ implies uniform continuity of $f^2$ on $\mathbb{R}$?

my question is if $f:\mathbb{R}\rightarrow\mathbb{R}$ is uniformly continuous, does it implies that $f^2$ is so?and in general even or odd power of that function?
Myshkin
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Is $f(x) = x/x$ the same as $f(x) = 1$?

Let $f(x) = \frac{x}{x}$. Is it correct to say that $f(x) \ne 1$, since $f(x)$ has a discontinuity at $x=0$?
Kenshin
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Continuity of an inverse function.

Theorem. Let $f\colon I \to J$, where $I$ is an interval and $J$ is the image $f(I)$, be a function such that: $f$ is strictly increasing on $I$. $f$ is continuous on $I$. Then $J$ is an interval, and $f$ has an inverse function $f^{-1}\colon J\to…
Charlie
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Points from continuous function form a square

Let a continuous function $f:[0,1]\mapsto\mathbb{R}$, such that $f(0)=f(1)=0$ and $f(x)>0$ for all $x\in (0,1)$. Prove or disprove that there are two points on the graph of $f$, such that with their projections to $x$-axis, form a square. I tried to…
STrick
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Lipschitz continuity of $f(x) \to x^2$

Is it legitimate to show the Lipshitz continuity of $f(x)=x^2$ as I did below? $$|x^2-y^2| \leq L|x-y| $$ w.l.o.g. $x>y.$ $$\Rightarrow x^2-y^2 \leq L|x-y|$$ $$\Leftrightarrow \ \dfrac {x^2-y^2}{x-y} \leq L\\ \Leftrightarrow \ \dfrac {(x-y) \cdot…
optional
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Continuity of $\sin(x)/x$

Is $\sin(x)/x$ continuous at $x=0$,(perfectly, not tends to)? The function has right hand and left hand limit same and equals 1 but what happens at perfect zero?it should be undefined. Is it?
user157588
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"Almost" continuous = continuous?

Let $x_0\in[a,b]$ be fixed and let $f:[a,b]\to\mathbb R$ be a function with the following property: For every $\varepsilon>0$ there exists some $\delta>0$ and some set $U\subset[x_0-\delta,x_0+\delta]$, where $[x_0-\delta,x_0+\delta]\backslash U$ is…
sranthrop
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Removable discontinuity or asymptote?

The difference between a "removable discontinuity" and a "vertical asymptote" is that we have a R. discontinuity if the term that makes the denominator of a rational function equal zero for x = a cancels out under the assumption that x is not equal…
Tim
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A trivial question about continuity

$f:\mathbb{R}\longrightarrow\mathbb{R}:x\mapsto x^2$ This function is continuous as we all know. Since for every point in the domain, we will always be able to draw a $\delta\epsilon-$rectangle, for every $\epsilon$ which captures every point of…
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Continuity at the origin of $f(x,y)$ given by $0$ there, and $\frac{\sqrt{|x||y|}}{|x|+|y|}$ elsewhere

Define the function $f:\mathbb{R}^2\mapsto \mathbb{R}$ by \begin{equation*} f(x,y) := \begin{cases} \frac{\sqrt{|x||y|}}{|x|+|y|} & \textit{if} \; \; (x,y)\neq (0,0) \\ 0 & \textit{if} \; \; (x,y) = (0,0) \end{cases}. \end{equation*} Determine…
squenshl
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