Continuity of a function as in real analysis The usual $\varepsilon-\delta$ definition of a continuous function is -
$\mathrm{Definition : }$ $~ L e t ~ $(X, d) and $\left(Y, d^{\prime}\right)$ be two metric spaces. A function $f: X \rightarrow Y$ is said to be continuous at $a \in X$ if for each   $\varepsilon&gt;0$ , there exists $\delta&gt;0$ such that whenever $x \in X$ satisfies $d(x, a)$ $&lt;\delta,$ we have $d^{\prime}(f(x), f(a))&lt;\varepsilon$.
Consider the function $f(x): X \rightarrow Y$ with $X$ and $Y$  two open disks, of radius $R$, in $2$D Euclidean space. $a$ is the centre of $X$ and $f(a)$ of $Y$. $f$ has the property, apart from  mapping the centre of $X$ to the centre of $Y$, that it maps a circular ring (of radius $r1$) in $X$ to a ring in $Y$ with radius $r2$ such that $r2 \rightarrow R$ when $r1 \rightarrow 0$ and $r2 \rightarrow 0$ when $r1 \rightarrow R$. In other words, $f$ maps a ring in $X$, with radius going from $R$ to $0$, to a ring in $Y$ with radius going from $0$ to $R$ ($R$ and $0$ obviously not included).

My question - The function satisfies the condition for continuity at $a$ with $\delta = R$, but $f(x)$ moves 'discontinuously' in $Y$ as we move slightly away from $a$ in $X$. Do we say the function is continuous at $a$ ?
Please correct me if there is any flaw in the thought process.
The function is discontinuous at $a$: Consider $\varepsilon&gt;0$ with $\varepsilon &lt; R$. Then no matter how small we choose $\delta&gt;0,$ any $\delta$-ball around $a$ will include some $x$ lying on a circle of sufficiently small radius $r_1$ for which $f(x)$ lies on a circle of radius $r_2&gt;\varepsilon$.
(Let me know if I should elaborate on this answer)
Short answer: no.  It is not the case that once you pick an $\varepsilon \ll R$ that all the points of $X$ are mapped within $\varepsilon$ of $f(a)$.  $a$ is mapped to $f(a)$ and the rest of a small disk around $a$ are mapped to an annulus with inner radius $\gg \varepsilon$.  That is, there are points within $R$ of $a$ that are not mapped within $\varepsilon$ of $f(a)$.  Any one of these points is a witness to failure of the implication in the definition of continuity.

I note that you make no requirement that the mapping of rings be continuous in terms of the radius.  You specify that a small enough annulus in $X$ is sent to a large enough annulus in $Y$ and vice versa, but make no constraint on radii in the &quot;middle&quot;.  As currently specified, these could be randomly shuffled entirely discontinuously.  In such a case, one can inspect the function $r_2(r_1)$, the image radius of each domain ring, for discontinuities.  But even if this function has no discontinuities (and even if the individual rings are oriented and distorted to give continuity on $X \smallsetminus \{a\}$)...

The obvious place to look for a discontinuity is in a neighborhood of the center(s) since the open ball of small radius is sent to the center of the image disk and to an annulus on the edge of the image disk.
It is a bit of a cheat that you have used &quot;$a$&quot; to be both the generic point in the definition of continuity and the center of one disk.  For continuity, the definition has to hold for every point on the disk, not just the center.  Nevertheless let $a = a$.  (That is, let us inspect the continuity around the center of $X$.)
Consider $r_2(r_1)$, the function representing the image ring radius in $Y$ of the domain ring in $X$ of radius $r_1$.  As per the discussion above the horizontal rule, we have already allowed this function to be continuous, so this is a continuous, strictly monotonically decreasing function of $r_1$.  (To see this, note injectivity, then apply the intermediate value theorem.)  This means it intersects the line $1 \cdot r_1 + 0$ exactly once.  That intersection corresponds to the one ring whose domain radius equals its image radius.  Call that radius, $r$.
Let $\varepsilon \in (0, r)$ and let $U$ be the disk in $X$, centered at $a$, of radius $\varepsilon$.  The image of $U$ is $\{f(a)\}$ union an annulus of inner radius greater than $r$.  Call that annulus $V$.
Now, make any choice of $\delta$ (including your proposed choice, $\delta = R$).  Then there is a point, $b$, in the disk of radius $\delta$ centered at $a$ that is also in $U \smallsetminus \{a\}$.  (We have intersected two open disks centered at the same point.  This gives an open disk with the smaller radius, which necessarily contains at least one point that is not the center.)  We have $d(a,b) &lt; \delta$.  Of course, $f(b)$ is in $V$, so $d(a,f(b) &gt;r &gt; \varepsilon$, so we have found a witness to discontinuity.