Continuity Carries Limits to Their Images
A sequence can converge without being described by distances: in any topological space, convergence means that the sequence is eventually in every open neighborhood of its proposed limit. Continuity interacts directly with this definition. If a sequence approaches \(x\) and a map is continuous, then its image sequence approaches \(f(x)\). The statement applies even when the spaces are not metric spaces and even when a sequence can have more than one limit.
Throughout, \((X,\mathcal{T})\) and \((Y,\mathcal{U})\) are topological spaces. A sequence \((x_n)\) in \(X\) converges to \(x\in X\) if, for every open set \(V\in\mathcal{T}\) containing \(x\), there is \(N\in\mathbb{N}\) such that \(x_n\in V\) for every \(n\ge N\). A map \(f:X\to Y\) is continuous if \(f^{-1}(W)\) is open in \(X\) for every open set \(W\) in \(Y\).
Proof. Let \(W\) be any open neighborhood of \(f(x)\) in \(Y\). Continuity makes \(f^{-1}(W)\) open in \(X\), and \(x\in f^{-1}(W)\) because \(f(x)\in W\). Since \(x_n\to x\), there is \(N\in\mathbb{N}\) such that \(x_n\in f^{-1}(W)\) for every \(n\ge N\). By the definition of preimage, this means \(f(x_n)\in W\) for every \(n\ge N\). Thus every open neighborhood of \(f(x)\) eventually contains the image sequence, so \(f(x_n)\to f(x)\). \(\square\)
The proof uses only open sets, preimages, and the definition of sequence convergence. It does not require a metric, nor does it require the limit to be unique. In particular, if \(x_n\) has two limits in \(X\), applying the theorem to each one gives corresponding limits of the same image sequence.
A Metric Example: Squaring a Convergent Sequence
Worked Example: The Square of a Sequence in the Real Line
Let \(X=Y=\mathbb{R}\), each with its standard topology, let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), and set \(x_n=2+1/n\) for \(n\in\mathbb{N}\). We verify the input limit, calculate the image sequence, and check its limit directly.
First, \(x_n\to 2\). Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). For every \(n\ge N\), \[ |x_n-2|=\left|2+\frac1n-2\right|=\frac1n\le\frac1N<\varepsilon. \] This is the metric definition of convergence.
For every \(n\in\mathbb{N}\), direct expansion gives \[ f(x_n)=\left(2+\frac1n\right)^2 =4+\frac4n+\frac1{n^2}. \] For \(n\ge1\), \(1/n^2\le1/n\), so \[ |f(x_n)-4|=\frac4n+\frac1{n^2}\le\frac5n. \] Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>5/\varepsilon\). Then \(n\ge N\) implies \(|f(x_n)-4|\le5/n\le5/N<\varepsilon\). Therefore \(f(x_n)\to4=f(2)\).
Check: the limit predicted by the theorem is \(f(2)=2^2=4\), and the explicit estimate shows that the difference from \(4\) tends to zero. The calculation also distinguishes the input limit \(2\) from the output limit \(4\).
For this application, \(f\) is continuous in the standard topology: for any \(c\in\mathbb{R}\) and any \(\varepsilon>0\), the identity \(|y^2-c^2|=|y-c||y+c|\) can be used to find a neighborhood of \(c\) whose points have squares within \(\varepsilon\) of \(c^2\). The general theorem itself does not depend on this particular way of establishing continuity.
A Metric Space Mapping to a Non-Hausdorff Space
To see how the theorem works when the target is not a metric space, consider the two-point space \(S=\{a,b\}\) with topology \(\{\varnothing,\{a\},S\}\). This is a topology: it contains the empty set and the whole space, and unions and finite intersections of its three sets remain among those sets. The open neighborhoods of \(a\) are \(\{a\}\) and \(S\); the only open neighborhood of \(b\) is \(S\).
Worked Example: A Continuous Map into the Sierpinski Space
Give \(\mathbb{R}\) its standard topology and let \(S\) have the topology just defined. Define \(f:\mathbb{R}\to S\) by \(f(x)=a\) for \(x<0\), and \(f(x)=b\) for \(x\ge0\). We first check continuity. The inverse images of the open sets in \(S\) are \[ f^{-1}(\varnothing)=\varnothing,\qquad f^{-1}(\{a\})=(-\infty,0),\qquad f^{-1}(S)=\mathbb{R}. \] All three are open in the standard topology of \(\mathbb{R}\). Since these are every open set in \(S\), \(f\) is continuous.
Now let \(x_n=-1/n\) for \(n\in\mathbb{N}\). For every \(\varepsilon>0\), choosing \(N>1/\varepsilon\) ensures that \(n\ge N\) gives \(|x_n-0|=1/n\le1/N<\varepsilon\). Thus \(x_n\to0\) in \(\mathbb{R}\). Since \(-1/n<0\) for every \(n\), \(f(x_n)=a\) for every \(n\), whereas \(f(0)=b\).
The image sequence converges to \(b\): the only open neighborhood of \(b\) is \(S\), and every term \(f(x_n)=a\) belongs to \(S\). It also converges to \(a\): each open neighborhood of \(a\), namely \(\{a\}\) or \(S\), contains every term. In particular, \(f(x_n)\to f(0)\), as the theorem requires, though this image sequence has another limit as well.
Check: \(f(x_n)=a\) for all \(n\), and the neighborhood test at \(b=f(0)\) asks only whether the sequence is eventually in \(S\). It is in \(S\) at every index.
This example makes an important distinction: the theorem says that the image sequence converges to the image of the chosen limit. It does not say that this is the image sequence’s only limit. The target space \(S\) is not Hausdorff: the points \(a\) and \(b\) cannot have disjoint open neighborhoods, since every neighborhood of \(b\) is \(S\).
Why Continuity and Hausdorffness Matter
Worked Example: A Discontinuous Map Can Lose a Limit
Let \(S=\{a,b\}\) have topology \(\{\varnothing,\{a\},S\}\), and let \(Y=\{0,1\}\) have the subspace topology inherited from standard \(\mathbb{R}\). The sets \(\{0\}\) and \(\{1\}\) are open in \(Y\): they are the intersections of \(Y\) with the open intervals \((-1/2,1/2)\) and \((1/2,3/2)\), respectively. Thus each singleton is an open neighborhood of its point.
Define \(g:S\to Y\) by \(g(a)=0\) and \(g(b)=1\). Let \(x_n=a\) for every \(n\in\mathbb{N}\). This sequence converges to \(b\) in \(S\), because the only open neighborhood of \(b\) is \(S\), which contains every term. But \(g(x_n)=0\) for every \(n\), and this image sequence does not converge to \(g(b)=1\): the open neighborhood \(\{1\}\) of \(1\) contains none of its terms.
The failure is consistent with the theorem because \(g\) is not continuous. In fact, \(g^{-1}(\{1\})=\{b\}\), and \(\{b\}\) is not open in \(S\). The calculation identifies exactly why continuity is needed: an open neighborhood of the proposed image limit need not have an open preimage neighborhood around the input limit.
Check: the input sequence is constantly \(a\) and converges to \(b\); the output sequence is constantly \(0\), which is not eventually in the neighborhood \(\{1\}\) of \(g(b)\).
There is a useful contrast with Hausdorff target spaces. A space is Hausdorff if any two distinct points have disjoint open neighborhoods. In such a space, a sequence cannot converge to two different points.
Proof. Suppose \(y\ne z\). By the Hausdorff property, there are disjoint open neighborhoods \(U\) of \(y\) and \(V\) of \(z\). Convergence to \(y\) gives an index \(N_1\) such that \(y_n\in U\) for every \(n\ge N_1\). Convergence to \(z\) gives an index \(N_2\) such that \(y_n\in V\) for every \(n\ge N_2\). Choose \(n\ge\max(N_1,N_2)\). Then \(y_n\in U\cap V\), contradicting \(U\cap V=\varnothing\). Therefore \(y=z\). \(\square\)
Combining this proposition with the preservation theorem gives a practical conclusion: if a sequence in \(X\) converges to \(x\), and the target \(Y\) is Hausdorff, then its image under a continuous map converges to \(f(x)\) and cannot converge to a different point. The Hausdorff assumption is needed for that uniqueness conclusion, not for preservation itself.
Composition and a Common Pitfall
Proof. By the continuous-images theorem applied to \(f\), \(f(x_n)\to f(x)\) in \(Y\). Applying the same theorem to \(g\) gives \(g(f(x_n))\to g(f(x))\) in \(Z\). These are exactly \((g\circ f)(x_n)\) and \((g\circ f)(x)\), respectively. \(\square\)
Common pitfall. The conclusion is not that the image sequence has the same limit value as the original sequence. In the first worked example, \(x_n\to2\), but the correct image limit is \(f(2)=4\), not \(2\). Indeed, \(f(x_n)=4+4/n+1/n^2\), and its difference from \(4\) is at most \(5/n\). The correction is to apply the map to the proposed input limit: for a continuous \(f\), the guaranteed output limit is \(f(x)\).
The result is useful whenever a sequence is easier to understand before applying a map. One can establish convergence in the domain and then transfer it to the codomain without separately analyzing every open neighborhood there. The proof also pinpoints the required hypothesis: continuity turns each open neighborhood of the image limit into an open neighborhood of the original limit. Without that step, even a sequence with a limit can have an image sequence that fails to converge to the image of that limit.
Check Your Understanding
Use the neighborhood definition of convergence and the stated topologies in each question.
- For a continuous map \(f:X\to Y\) and a sequence \(x_n\to x\), which open set in \(X\) is used to prove that \(f(x_n)\to f(x)\)?
- In the Sierpinski space \(S=\{a,b\}\) with topology \(\{\varnothing,\{a\},S\}\), does the constant sequence \(x_n=b\) converge to \(a\)? Justify your answer using the open neighborhoods of \(a\).
- For the map \(f:\mathbb{R}\to S\) defined by \(f(x)=a\) when \(x<0\) and \(f(x)=b\) when \(x\ge0\), find the image of \(x_n=-1/n\) and state all its limits in \(S\).
- Why is the map \(g:S\to\{0,1\}\) with \(g(a)=0\) and \(g(b)=1\) not continuous?
- Where does the Hausdorff assumption enter the proof that a convergent sequence has at most one limit?