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Can you find a map R to R that is open but not continuous?

以下の文章がところどころ意味がわからなくて、完全に理解したくてもなかなかできないでいます。この英文の主張を解説していただけないでしょうか?よろしくお願いします。 [原文] As R is homeomorphic to the open interval (0,1), we will define a function f from R to (0,1) that satisfies the requirements. You can then extend it to a function from R to R by composing it with, for example, h(x) = (2x-1)/(x(1-x)) or a similar function. We compute x in the following way: * First, drop the integer part of x - our function will be periodic. * Next, write x in decimal representation. If you have the choice of two representations, for example, 0.5 = 0.4999..., choose the first one. * Find the last digit '9' in the number (if there is no last 9, see below). * Remove all the digits up to and including that '9', and consider the remaining digits. As there will be no '9' among them, you can interpret them as the "decimal" expansion of a number in base 9 - that is your f(x), and it will be between 0 and 1. However, a few things may go wrong: * There may be no 9 at all in the number, or an infinite number of 9 (not necessarily contiguous). * The resulting base 9 number may be 0 or 0.888... = 1 (in base 9) In both of these cases, set f(x) = 0.5 (or any number between 0 and 1). For example, to get f(3.1952945): * Drop everything up to the last 9. * The remaining number is 0.45, which, interpreted as a number in base 9, is 41/81. We claim that the image of any open interval is the whole open interval (0,1), which is an open set. This will show that f is open. Indeed, given an interval I = (a,b), we can find a sub-interval J of length 10^(-k) contained in it, for a suitable k > 1, such that the interval fixes the first (k-1) digits after the decimal point. For example, the interval I = (1.14567, 1.15321) contains the sub-interval J = (1.151, 1.152). Given any y in (0,1), we can find an x in J (and therefore in I) such that f(x) = y. To do that: * write y in base 9 * append a digit 9 to the lower bound of J. * append the base 9 representation of y. For example, with I and J as above, we can find an x such that f(x) = 1/3. * 1/3 in base 9 is 0.3. * The lower bound of J is 1.151. * the number x = 1.15193 satisifies f(x) = y, and belongs to J. Note that there are many other possible numbers, obtained by choosing smaller intervals for J, for example, 1.1512393, and so on. It should also be obvious that f is not continuous (in fact it is in some sense the most discontinuous possible function). Indeed, a function f is continuous at a if, given any open interval K = (f(a)-e,f(a)+e), we can find an open interval J = (a-d,a+d) such that the image of J is a subset of K. But, by what we just saw, the image of any interval is the whole of (0,1).

noname#89074

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いや,こりゃなんでもひどすぎでしょう. もっと的を絞りなさいな. ちょっと前に, >* The remaining number is 0.45, which, interpreted as a number in base 9, is 41/81. のところだけ,聞いてたけどねぇ・・・ 数学のある程度(とはいってもきわめて初歩の数学) 専門的な英文だから,英語カテゴリに行っても 厳しいだろうが・・・これって, 丸ごと一個の問題というか塊全部でしょう? それにタイトルと中身が違うし。。。 >Can you find a map R to R that is open but not continuous? ↑これって意味がわかってますか? ちなみに・・・この文章は位相空間というよりは, 解析への応用を前提として, それに必要な位相の準備をしてるような 感じかな・・・.近くにカントールの三進集合とか シェルピンスキーの三角形の話があったりして. この文章は, 任意の開区間を(0,1)に写像する関数を 実際に構築してるってことですが, もっと前に何か文章があるのでしょう. 最初の二行とそのあとの「構築」のつながりが見えません. そもそも >we will define a function f from R to (0,1) that satisfies the requirements. の「the requirement」って何でしょう? ここで構築しているfは明らかにhomeoではないし・・・ こういう文章を読み解くには ただ文字を追うのではなく, 実際に具体例を計算しながら読むのです f(3.141592)じゃなくって,f(0.1)とかf(5.1)とか いろいろな開区間に対してJってどうなるとか. 自分で計算してみました? 計算してみれば英語を多少誤読してても その計算が誤読を修正してくれるものです

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質問者からの補足

回答ありがとうございます。 タイトルは「1次元ユークリッド空間RからRへの写像fで、開写像だが連続でないものを見つけれますか?」という意味だと解釈してます。 すいませんm(_ _)m 質問の仕方が雑でした。 本当に申し訳ありません。 自分の訳を一緒に載せ、再度投稿しなおそうと思います。

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  • この英文の数学的意味を教えてください。

    以下の英文を考えているのですが、よくわからない箇所があります。 ()で囲まれた箇所は私が自分で意訳したものです。 「 Find a map R to R that is open but not continuous」 (1次元ユークリッド空間RからRへの写像fで、開写像であるが、連続ではないものを見つけてください。) As R is homeomorphic to the open interval (0,1), we will define a function f from R to (0,1) that satisfies the requirements. You can then extend it to a function from R to R by composing it with, for example, h(x) = (2x-1)/(x(1-x)) or a similar function. (Rは開区間(0,1)と同相であるから、Rから(0,1)への写像fで、開写像だが連続でないものを見つければ十分です。実際、そのような写像fを見つけることが出来れば、(0,1)からRへの同相写像として例えば、 h(x) = (2x-1)/(x(1-x)) を考え、これとfとの合成写像を考えます。するとそれがRからRへの開写像で、かつ連続でないものだからです。) We compute x in the following way: (xを以下のように計算することにします。) * First, drop the integer part of x - our function will be periodic. (まず最初に、xの整数部分を0にします。) * Next, write x in decimal representation. If you have the choice of two representations, for example, 0.5 = 0.4999..., choose the first one. (次に、xを10進法表記します。もし、2通りの表記が存在する場合、たとえば0.5 = 0.4999・・・などの場合、前者を採用します。) * Find the last digit '9' in the number (if there is no last 9, see below). (小数部分の中にある"9"のうち、最も位の小さいところにある"9"を見つけます。(そのような9がない場合については後述)) * Remove all the digits up to and including that '9', and consider the remaining digits. As there will be no '9' among them, you can interpret them as the "decimal" expansion of a number in base 9 - that is your f(x), and it will be between 0 and 1. (そして、その9より前にある数すべてを取り除きます。ただし9自身も取り除く対象です。そして、取り除いた後の数を"9進数"表記だと解釈し、それをf(x)とします。するとそれはf(x)∈(0,1)です) However, a few things may go wrong: (しかしながら、すこしマズイ箇所があります。) * There may be no 9 at all in the number, or an infinite number of 9 (not necessarily contiguous). (整数部分を0にし、10進数で小数展開した数に9が現われなかったり、9が無限に現れる(ただし断続的でもよい)場合があるからです。) * The resulting base 9 number may be 0 or 0.888... = 1 (in base 9) (そのような数を9進数表記だと解釈すると、それは0か0,888・・・のどちらかでしょう。) In both of these cases, set f(x) = 0.5 (or any number between 0 and 1). (このどちらかのケースでは、f(x)を0.5と定義します((0,1)の元であれば何でも良いです。)) For example, to get f(3.1952945): * Drop everything up to the last 9. * The remaining number is 0.45, which, interpreted as a number in base 9, is 41/81. (例えば、f(3.1952945)を計算したい場合、3.1952945を0.31952945にし、さらに最後に現れる9より前を取り除き0.45にします。0.45を9進数表記されたものだと解釈すれば0,45は 41/81です。これが求める値です。) We claim that the image of any open interval is the whole open interval (0,1), which is an open set. This will show that f is open. (すると、どんな開区間のfによる像も、開区間(0,1)と一致します。今、(0,1)は開集合であるから、これはfが開写像であることに他ならない。) ※開写像である理由ですが、始集合Rにおける開基として特に、開区間全体をとれるので、任意の開区間のfによる像が、終集合Rにおける開集合となることさえチェックできれば十分だからです。 本当は、まだこれ以降も続きがあるのですが、とりあえずこの段階で分からないことがあるので質問させてください。 (1) The resulting base 9 number may be 0 or 0.888... = 1 (in base 9)の訳はあっているでしょうか?仮にそうだとして、なぜ0or0,88・・・なのでしょうか? (2)We claim that the image of any open interval is the whole open interval (0,1), which is an open set. This will show that f is open.の訳はあっているでしょうか? 仮にそうだとして、"一致"するのはなぜでしょうか? fの定義から(0,1)に含まれるのはわかるのですが、逆の包含関係が成り立つ理由がわからないです。 どなたか解説していただけないでしょうか?よろしくお願い致します。

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  • 回答No.1
noname#88823

英語は英語のカテゴリーで聞いてください。

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質問者からのお礼

すこし調べましたが、数学カテゴリーには 英語でのtopologyに関する質問はたくさん投稿されていますが・・・。 タイトルの書き方がまずかったですかね? これ以上回答者が現れない場合は、タイトルを日本語にして再投稿しようと思います。

質問者からの補足

回答ありがとうございます。 この英文は内容が位相空間論に関する話ですが、英語のカテゴリーで質問した場合、数学的意味を教えていただけるのでしょうか? 数学のカテゴリーならば、位相を洋書で学ばれた方もたくさんいるだろうと思い投稿しているのですが、そこらへんはどうなのでしょうか?

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